Rd Sharma Xi 2020 2021 _volume 2 Solutions for Class 11 Science Maths Chapter 30 Derivatives are provided here with simple step-by-step explanations. These solutions for Derivatives are extremely popular among Class 11 Science students for Maths Derivatives Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma Xi 2020 2021 _volume 2 Book of Class 11 Science Maths Chapter 30 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma Xi 2020 2021 _volume 2 Solutions. All Rd Sharma Xi 2020 2021 _volume 2 Solutions for class Class 11 Science Maths are prepared by experts and are 100% accurate.

Page No 30.25:

Question 1:

(i) 2x

(ii) 1x

(iii) 1x3

(iv) x2+1x

(v) x2-1x

(vi) x+1x+2

(vii) x+23x+5

(viii) k xn

(ix) 13-x

(x) x2 + x + 3

(xi) (x + 2)3

(xii) x3 + 4x2 + 3x + 2

(xiii) (x2 + 1) (x − 5)

(xiv) 2x2+1

(xv) 2x+3x-2

Answer:

i ddxf(x)=limh0fx+h-fxh                 =limh02x+h-2xh                 =limh02x-2x-2hhx(x+h)                 =limh0-2hhx(x+h)                 =limh0-2x(x+h)                 =-2x2
ii ddxf(x)=limh0fx+h-fxh                 =limh01x+h-1xh                 =limh0x-x+hhxx+h×x+x+hx+x+h                 =limh0x-x-hhxx+hx+x+h                 =limh0-hhxx+hx+x+h                 =limh0-1xx+hx+x+h                 =-1xxx+x                 =-1x× 2x                 =-12x32                 =-12x-32
iii ddxf(x)=limh0fx+h-fxh                  =limh01(x+h)3-1x3h                  =limh0x3-(x+h)3h(x+h)3x3                  =limh0x3-x3-3x2h-3xh2-h3h(x+h)3x3                  =limh0-3x2h-3xh2-h3h(x+h)3x3                  =limh0h-3x2-3xh-h2h(x+h)3x3                  =limh0-3x2-3xh-h2(x+h)3x3                  =-3x2x6                  =-3x4                  =-3x-4
iv ddxf(x)=limh0fx+h-fxh                     =limh0(x+h)2+1x+h-x2+1xh                     =limh0x2+2xh+h2+1x+h-x2+1xh                     =limh0x3+2x2h+h2x+x-x3-x2h-x-hxh(x+h)                     =limh0x2h+h2x-hx(x+h)                     =limh0h(x2+hx-1)xh(x+h)                     =limh0x2+hx-1x(x+h)                     =x2-1x2

v ddxf(x)=limh0fx+h-fxh                     =limh0(x+h)2-1x+h-x2-1xh                     =limh0x2+2xh+h2-1x+h-x2-1xh                     =limh0x3+2x2h+h2x-x-x3-x2h+x+hxh(x+h)                     =limh0x2h+h2x+hx(x+h)                     =limh0h(x2+hx+1)xh(x+h)                     =limh0x2+hx+1x(x+h)                     =x2+1x2

vi ddxf(x)=limh0fx+h-fxh                     =limh0x+h+1x+h+2-x+1x+2h                     =limh0x+h+1x+2-x+h+2x+1hx+h+2x+2                     =limh0x2+2x+hx+2h+x+2-x2-x-hx-h-2x-2hx+h+2x+2                     =limh0hhx+h+2x+2                     =limh01x+h+2x+2                     =1x+0+2x+2                     =1x+22

vii ddxf(x)=limh0fx+h-fxh                     =limh0x+h+23x+h+5-x+23x+5h                     =limh0x+h+23x+3h+5-x+23x+5h                     =limh0x+h+23x+5-3x+3h+5x+2h3x+3h+53x+5                     =limh03x2+3xh+6x+5x+5h+10-3x2-3xh-5x-6x-6h-10h3x+3h+53x+5                     =limh0-hh3x+3h+53x+5                     =limh0-13x+3h+53x+5                     =-13x+52

viii ddxf(x)=limh0fx+h-fxh                     =limh0 k x+hn-kxnh                     =limx+h-x0 k x+hn-xnx+h-xHere, we have: limxaxm-amx-a=m am-1                     =k n xn-1

ix ddxf(x)=limh0fx+h-fxh                     =limh0 13-x-h-13-xh                     =limh0 3-x-3-x-hh3-x3-x-h                     =limh0 3-x-3-x-hh3-x3-x-h× 3-x+3-x-h3-x+3-x-h                     =limh0 3-x-3+x+hh3-x3-x-h3-x+3-x-h                     =limh0 hh3-x3-x-h3-x+3-x-h                     =limh0 13-x3-x-h3-x+3-x-h                     =13-x3-x-03-x+3-x-0                     =13-x 23-x                     =123-x32

x ddxf(x)=limh0fx+h-fxh                     =limh0 x+h2+x+h+3-x2+x+3h                     =limh0x2+h2+2xh+x+h+3-x2-x-3h                     =limh0h2+2xh+hh                     =limh0 h(h+2x+1)h                     =limh0 h+2x+1                     =0+2x+1                     =2x+1

xi ddxf(x)=limh0fx+h-fxh                     =limh0 x+h+23-x+23h                     =limh0x+h+2-x-2x+h+22+x+h+2x+2+x+22h                     =limh0hx+h+22+x+h+2x+2+x+22h                     =limh0 x+h+22+x+h+2x+2+x+22                     =x+0+22+x+0+2x+2+x+22                     =x+22+x+22+x+22                     =3x+22

xii ddxf(x)=limh0fx+h-fxh                     =limh0 x+h3+4x+h2+3x+h+2-x3+4x2+3x+2h                     =limh0x3+3x2h+3xh2+h3+4x2+4h2+8xh+3x+3h+2-x3-4x2-3x-2h                     =limh0 3x2h+3xh2+h3+4h2+8xh+3h+2h                     =limh0 h3x2+3xh+h2+4h+8x+3h                     =limh03x2+3xh+h2+4h+8x+3                     =3x2+8x+3

xiii x2+1x-5=x3-5x2+x-5ddxf(x)=limh0fx+h-fxh                     =limh0 x+h3-5x+h2+x+h-5-x3-5x2+x-5h                     =limh0 x3+3x2h+3xh2+h3-5x2-5h2-10xh+x+h-5-x3+5x2-x+5h                     =limh0 3x2h+3xh2+h3-5h2-10xh+hh                     =limh0 h 3x2+3xh+h2-5h-10x+1h                     =limh0 3x2+3xh+h2-5h-10x+1                     =3x2-10x+1

xiv ddxf(x)=limh0fx+h-fxh                     =limh0 2x+h2+1-2x2+1h                     =limh02x2+2h2+4xh+1-2x2+1h                     =limh02x2+2h2+4xh+1-2x2+1h×2x2+2h2+4xh+1+2x2+12x2+2h2+4xh+1+2x2+1                     =limh0 2x2+2h2+4xh+1-2x2-1h2x2+2h2+4xh+1+2x2+1                     =limh0 h2h+4xh2x2+2h2+4xh+1+2x2+1                     =limh0 2h+4x2x2+2h2+4xh+1+2x2+1                     =4x2x2+1+2x2+1                     =4x22x2+1                     =2x2x2+1

xv ddxf(x)=limh0fx+h-fxh                     =limh0 2x+h+3x+h-2-2x+3x-2h                     =limh0 2x+2h+3x-2-x+h-22x+3hx+h-2x-2                     =limh0 2x2+2xh+3x-4x-4h-6-2x2-2xh+4x-3x-3h+6hx+h-2x-2                     =limh0 -7hhx+h-2x-2                     =limh0 -7x+h-2x-2                     =-7x-2x-2                     =-7x-22

Page No 30.25:

Question 2:

Differentiate each of the following from first principles:

(i) ex

(ii) e3x

(iii) eax + b

(iv) x ex

(v) − x

(vi) (−x)−1

(vii) sin (x + 1)

(viii) cosx-π8

(ix) x sin x

(x) x cos x

(xi) sin (2x − 3)

Answer:

i ddxf(x)=limh0fx+h-fxhddxex=limh0e-(x+h)-e-xh             =limh0e-xe-h-e-xh             =limh0e-xe-h-1h             =-e-xlimh0e-h-1-h             =-e-x1             =-e-x

ii ddxf(x)=limh0fx+h-fxhddxe3x=limh0e3(x+h)-e3xh             =limh0e3xe3h-e3xh             =limh0e3xe3h-13h             =3 e3xlimh0e3h-13h             =3 e3x1             =3 e3x

iii ddxf(x)=limh0fx+h-fxhddxeax+b=limh0ea(x+h)+b-eax+bh             =limh0eax+beah-eax+bh             =limh0eax+beah-1h             =a eax+blimh0eah-1ah             =a eax+b1             =a eax+b

iv ddxf(x)=limh0fx+h-fxhddxx ex=limh0(x+h )e(x+h)-x exh                =limh0(x+h) exeh-x exh                =limh0x exeh+hexeh-x exh                =limh0x exeh-x exh+limh0h exehh                =limh0x exeh-1h+limh0exeh                =xex1+exe0                =xex+ex

v ddxfx=limh0fx+h-fxhddx-x=limh0-x+h--xh              =limh0-x-h+xh              =limh0-hh              =limh0-1              =-1

vi -x-1=1-x ddxfx=limh0fx+h-fxhddx1-x =limh01-x+h-1-x  h                =limh0-1x+h+1xh                =limh0-x+x+hh x x+h                =limh0hh x x+h                =limh01 x x+h                =1x.x                =1x2

vii ddxfx=limh0fx+h-fxhddxsin x+1=limh0sin x+h+1-sin x+1 hWe know:sin C-sin D=2 cos C+D2 sin C-D2                       =limh02 cos x+h+1+x+12 sin x+h+1-x-12h                       =limh02 cos 2x+h+22 sin h2 h                       =2limh0 cos 2x+h+22   limh0 sin h2 h2×12                       =2 cos x+1 ×12                       =cos x+1

viii ddxfx=limh0fx+h-fxhddxcos x-π8=limh0cos x+h-π8-cos x-π8hWe know:cos C-cos D=-2 sin C+D2 sin C-D2                        =limh0-2 sin x+h-π8+x-π82 sin x+h-π8-x+π82h                        =limh0-2 sin 2x+h-π42 sin h2 h                        =-2limh0 sin 2x+h-π42 limh0 sin h2 h2×12                        =-2 sin x-π8 ×12                        =-sin x-π8

ix ddxf(x)=limh0fx+h-fxh                  =limh0 x+h sinx+h - x sin xh                  =limh0x+hsin x cos h + cos x sin h- x sin xh                  =limh0 x sin x cos h + x cos x sin h +h sin x cos h + h cos x sin hh                  =limh0 x sin x cos h  - x sin x + x cos x sin h +h sin x cos h + h cos x sin h - x sin x h                  =x sin x limh0   cos h -1h+x cos x limh0 sin h h+sin xlimh0 cos h+cos xlimh0 sin h                  =x sin x limh0 -2 sin2 h2h24×h4+ x cos x 1+ sin x 1+ cos x 0                 = x sin x × -h2 + x cos x 1+ sin x 1+ cos x 0                  =-2x sin x 120+ x cos x + sin x                   = x cos x + sin x  

x ddxf(x)=limh0fx+h-fxh                   =limh0 x+h cos x+h - x cos xh                   =limh0x+hcos x cos h-sin x sin h- x cos xh                   =limh0 x cos x cos h - x sin x sin h+h cos x cos h - h sin x sin h- x cos xh                   =limh0 x cos x cos h  - x cos x - x sin x sin h+h cos x cos h - h sin x sin hh                   =x cos x limh0   cos h -1h-x sin x limh0 sin h h+cos xlimh0 cos h+sin xlimh0 sin h                   =x cos x limh0 -2 sin2 h2h24×h4-x sin x 1+cos x 1+ sin  x 0                  =xcosx  lim h0-h2-x sin x 1+cos x 1+ sin  x 0                  =-x cos x 0-x sin x+ cos x                    = -x sin x+ cos x  

xi ddxf(x)=limh0fx+h-fxh                    =limh0 sin 2x+2h-3-sin 2x-3hWe know:sin C-sin D=2 cos C+D2 sinC-D2                     =limh0 2 cos 2x+2h-3+2x-32 sin 2x+2h-3+2x-32 h                    =limh0 2 cos 4x+2h-62 sin h h                    =limh0 2 cos 4x+2h-62 limh0 sin hh                    =2 cos 4x-62 1                    = 2 cos 2x-3 



Page No 30.26:

Question 3:

Differentiate each of the following from first principles:

(i) sin 2x

(ii) sin xx

(iii) cos xx

(iv) x2 sin x

(v) sin (3x+1)

(vi) sin x + cos x

(vii) x2 ex

(viii) ex2+1

(ix) e2x

(x) eax+b

(xi) ax

(x) 3x2

Answer:

i ddxf(x)=limh0fx+h-fxh                   =limh0 sin 2x+2h-sin 2xh×sin 2x+2h+sin 2xsin 2x+2h+sin 2x                   =limh0sin 2x+2h- sin 2xh sin 2x+2h+sin 2xWe have:sin C-sin D= 2 cos C+D2 sin C-D2                   =limh02 cos 2x+2h+2x2 sin 2x+2h-2x2h sin 2x+2h+sin 2x                   =limh02 cos 2x+h sin hh sin 2x+2h+sin 2x                   =limh0 2 cos 2x+h   limh0 sin h hlimh01sin 2x+2h+sin 2x                    = 2 cos 2x 1 1sin 2x+sin 2x                   =2 cos 2x2sin 2x                   = cos 2xsin 2x

ii ddxf(x)=limh0fx+h-fxh                    =limh0 sin x+hx+h-sin xxh                    =limh0 x sin x+h- x+h sin x h x x+h                    =limh0 x sin x cos h + cos x sin h- x sin x - h sin xh x x+h                    =limh0 x sin x cos h +x cos x sin h- x sin x - h sin xh x x+h                    =limh0 x sin x cos h- x sin x +x cos x sin h - h sin xh x x+h                    =x sin xlimh0cos h -1h+x cos x x limh0 sin hhlimh0 1x+h-sin xxlimh01x+h                     =x sin x limh0 -2 sin2 h2h+x cos x x limh0 sin hhlimh0 1x+h-sin xxlimh01x+h                     =x sin x limh0 -2 sin2 h2h24×h4+x cos x x limh0 sin hhlimh0 1x+h-sin xxlimh01x+h                    =-x sin x ×limh0h2+x cos x x limh0 sin hhlimh0 1x+h-sin xxlimh01x+h                    =-x sin x 12 0+cos xx-sin xx2                    =cos xx-sin xx2                    =x cos x-sin xx2 

iii ddxf(x)=limh0fx+h-fxh                   =limh0 cos x+hx+h-cos xxh                   =limh0 x cos x+h- x+h cos x h x x+h                   =limh0 x cos x cos h - sin x sin h- x cos x - h cos xh x x+h                   =limh0 x cos x cos h -x sin x sin h- x cos x - h cos xh x x+h                   =limh0 x cos x cos h - x cos x-x sin x sin h - h cos xh x x+h                   =xcos xlimh0cos h -1h-xsin x x limh0 sin hhlimh0 1x+h-cos xxlimh01x+h                    =x cos x limh0 -2 sin2 h2h24×h4-xsin x x limh0 sin hhlimh0 1x+h-cos xxlimh01x+h                  limh0 sin2 h2h24=limh0sin h2h2×limh0sin h2h2=1×1, i.e. 1                    = -x cosxlimh0 h2-xsin x x limh0 sin hhlimh0 1x+h-cos xxlimh01x+h                  =-x cos x ×0-sin x 11x-cos x x1x                   =0-sin xx-cos xx2                   =-sin xx-cos xx2                   =-x sin x-cos xx2 

iv ddxf(x)=limh0fx+h-fxh                      =limh0 x+h2 sin x+h-x2 sin xh                      =limh0 x2+h2+2xhsin x cos h + cos x sin h-x2 sin xh                      =limh0 x2 sin x cos h + x2 cos x sin h + h2 sin x cos h +h2 cos x sin h+2xh sin x cos h +2xh cos x sin h-x2 sin xh                      =limh0x2 sin x cos h-x2 sin x + x2 cos x sin h + h2 sin x cos h +h2 cos x sin h+2xh sin x cos h +2xh cos x sin hh                      =x2 sin xlimh0cos h -1h+x2 cos xlimh0sin hh+sin x limh0 h cos h+ cos x limh0 h sin h +2x sin xlimh0 cosh +2x cos x limh0 sin h                      =x2 sin x limh0 -2 sin2 h2h24×h4+x2 cos xlimh0sin hh+sin x limh0 h cos h+ cos x limh0 h sin h +2x sin xlimh0 cosh +2x cos x limh0 sin h          limh0 sin2 h2h24=limh0sin h2h2×limh0sin h2h2=1×1, i.e. 1                      =-x2sinx× limh0 h2+x2 cos xlimh0sin hh+sin x limh0 h cos h+ cos x limh0 h sin h +2x sin xlimh0 cosh +2x cos x limh0 sin h                             =-x2 sin x ×0+x2 cos x 1+ sin x 0+ cos x 0+ 2x sin x 1+2x cos x 0                      =0+x2 cos x+ 2x sin x                      =0+x2 cos x+ 2x sin x                      =x2 cos x+ 2x sin x 

v ddxf(x)=limh0fx+h-fxh                        = limh0 sin 3x+h+1-sin 3x+1h                                         =limh0 sin 3x+3h+1-sin 3x+1h×sin 3x+3h+1+sin 3x+1sin 3x+3h+1+sin 3x+1                      =limh0sin 3x+3h+1-sin 3x+1h sin 3x+3h+1+sin 3x+1We have:  sin C-sin D= 2 cos C+D2 sin C-D2                      =limh02 cos 3x+3h+1+3x+12 sin 3x+3h+1-3x-12h sin 3x+3h+1+sin 3x+1                      =limh02 cos 6x+3h+22 sin 3h2h sin 3x+3h+1+sin 3x+1                      =limh0 2 cos 6x+3h+22   limh0 sin 3h 2h×32×32×limh01sin 3x+3h+1+sin 3x+1                       = 2 cos 3x+1 ×32 ×1sin 3x+1+sin 3x+1                      =3 cos  3x+12sin 3x+1 

vi ddxf(x)=limh0fx+h-fxh                    =limh0 sin x+h+cos x+h-sin x-cos xh                    =limh0sin x+h -sin xh+limh0 cos x+h-cos xhWe have: sin C-sin D= 2 cos C+D2 sin C-D2And, cos C-cos D=-2 sin C+D2 sin C-D2                    =limh0 2 cos 2x+h2 sin h2h+limh0 -2 sin 2x+h2 sin h2h                    =2 limh0 cos 2x+h2 limh0 sin h2h2×12 -2 limh0 sin 2x+h2  limh0 sin h2h2×12                    =2 cos x× 12-2 sin x ×12                    =cos x - sin x 

vii ddxf(x)=limh0fx+h-fxhddxx2 ex=limh0(x+h)2e(x+h)-x2exh                =limh0(x2+2xh+h2)exeh-x2exh                =limh0x2exeh+2xhexeh+h2exeh-x2exh                =limh0x2exeh-x2exh+limh02 x h exehh+limh0h2exehh                =limh0x2exeh-1h+limh02 x exeh+limh0 h exeh                =x2ex1+2xex1+0                =x2ex+2xex                =x2+2x ex

viii ddxf(x)=limh0fx+h-fxhddxex2+1=limh0e(x+h)2+1-ex2+1h                   =limh0ex2+h2+2xh+1-ex2+1h                   =limh0ex2+1eh2+2xh-ex2+1h                   =limh0ex2+1ehh+2x-1h×h+2xh+2x                   =ex2+1limh0 ehh+2x-1hh+2x limh0 h+2x                   =ex2+11 2x                   =2x ex2+1

ix ddxf(x)=limh0fx+h-fxhddxe2x=limh0e2(x+h)-e2xh                   =2 limh0e2x+2h-e2x2x+2h-2x                   =2 limh0 e2xe2x+2h-2x-12x+2h2-2x2                   =2 e2x limh0 e2x+2h-2x-12x+2h-2x2x+2h+2x                   =2 e2x limh0 e2x+2h-2x-12x+2h-2x limh012x+2h+2x                   =2 e2x 1122x                   =e2x2x

x ddxf(x)=limh0fx+h-fxhddxeax+b=limh0eax+ah+b-eax+bh                        =a limh0eax+ah+b-eax+bax+ah+b-ax+b                        =a limh0eax+beax+ah+b-ax+b-1ax+ah+b2-ax+b2                        =a eax+b limh0 eax+ah+b-ax+b-1ax+ah+b-ax+bax+ah+b+ax+b                        =a eax+b limh0 eax+ah+b-ax+b-1ax+ah+b-ax+b limh01ax+ah+b+ax+b                        =a eax+b 112ax+b                        =a eax+b2ax+b

xi ddxf(x)=limh0fx+h-fxhddxax=limh0ax+h-axh                 =limh0axax+h-x-1x+h-x                  =axlimh0ax+h-x-1x+h2-x2                 =axlimh0ax+h-x-1x+h-xx+h+x                 =axlimh0ax+h-x-1x+h-x limh01x+h+x                 =ax loge a 12x                 =12xax loge a

xii ddxf(x)=limh0fx+h-fxhddx3x2=limh03x+h2-3x2h              =limh0 3x2+2xh+h2-3x2h             =limh0 3x2 3x2+2xh+h2-x2-1h×h+2xh+2x              =3x2 limh0 3hh+2x-1hh+2xlimh0h+2x              =3x2 log 3 2x              =2x 3x2 log 3

Page No 30.26:

Question 4:

(i) tan2 x

(ii) tan (2x + 1)

(iii) tan 2x

(iv) tan x

Answer:

i ddxf(x)=limh0fx+h-fxh                =limh0tan2 x+h-tan2 xh                =limh0 tan x+h+tan xtan x+h-tan xh                =limh0 sin x+hcos x+h+sin xcos xsin (x+h)cos (x+h)-sin xcos x h                =limh0 sin x+h cos x + cos x+h sin x sin x+h cos x - cos x+h sin x h cos2 x cos2 x+h                 =limh0 sin 2x+hsin hh cos2 x cos2 x+h                 =1cos2 xlimh0 sin 2x+h  limh0 sin hh limh0 1cos2x+h                =1cos2 x sin 2x 11cos2 x                =1cos2 x 2 sin x cos x 1cos2 x                =2× sin x cos x×1cos2 x                =2 tan x sec2x

ii ddxf(x)=limh0fx+h-fxh                   =limh0 tan 2x+2h+1-tan 2x+1h                   =limh0sin 2x+2h+1cos 2x+2h+1-sin 2x+1cos 2x+1h                   =limh0 sin 2x+2h+1 cos 2x+1-cos 2x+2h+1 sin 2x+1 h cos 2x+2h+1 cos 2x+1                   =limh0 sin 2x+2h+1-2x-1h cos 2x+2h+1 cos 2x+1                   =1cos 2x+1limh0 sin 2h2h×2 limh01 cos 2x+2h+1                   =1cos 2x+1 ×2×1cos 2x+1                   =2cos22x+1                   =2 sec2 2x+1

iii ddxf(x)=limh0fx+h-fxh                    =limh0 tan 2x+2h-tan 2xh                   =limh0sin 2x+2hcos 2x+2h-sin 2xcos 2xh                   =limh0 sin 2x+2h cos 2x-cos 2x+2h sin 2x h cos 2x+2h cos 2x                   =limh0 sin 2x+2h-2xh cos 2x+2h cos 2x                   =1cos 2xlimh0 sin 2h 2h×2× limh01 cos 2x+2h                   =1cos 2x×2×1cos 2x                   =2cos22x                   =2 sec2 2x

iv ddxf(x)=limh0fx+h-fxh                    =limh0 tanx+h-tan xh×tanx+h+tan xtanx+h+tan x                   =limh0tanx+h-tan xhtanx+h+tan x                   =limh0sin x+hcos x+h-sin xcos xhtanx+h+tan x                   =limh0 sin x+h cos x-cos(x+h) sin x htanx+h+tan x cos x+h cos x                    =limh0 sin hhtanx+h+tan x cos x+h cos x                                =limh0 sin hhlimh0 1tanx+h+tan x cos x+h cos x                   =112 tan xcos2 x                   =sec2 x2 tan x

Page No 30.26:

Question 5:

(i) sin 2x
(ii) cos x
(iii) tan x
(iv) tan x2

Answer:

i Let f(x)= sin2x Thus, we have: f(x+h)=sin2x+hddxf(x)=limh0fx+h-fxh=limh0 sin 2x+2h-sin 2xhWe know:sin C- sin D=2 sin C-D2 cos C+D2=limh0 2 sin2x+2h-2x cos2x+2h-2x h=limh0 2×2 sin2x+2h-2x2 cos2x+2h+2x2 2h+2x-2x=limh0 2×2 sin2x+2h-2x2 cos2x+2h-2x2 2x+2h-2x2x+2h+2x=limh0 2×2 sin2x+2h-2x2  cos2x+2h-2x2 2× 2x+2h-2x22x+2h+2x= limh0 sin2x+2h-2x22x+2h-2x2limh0  2cos 2x+2h-2x22x+2h+2x   =1× 2cos2x22x                  limh0sin2x+2h-2x22x+2h-2x2=1=cos2x2x


(ii)  Let f(x) = cos x Thus, we have: f(x+h)=cos x+hddxfx=limh0f(x+h)-f(x) h= limh0cos x+h-cos xhWe know: cos C -cos D = -2sinC+D2 sinC-D2= limh0 -2sin x+h+x2 sinx+h-x2h= limh0 -2sin x+h+x2 sinx+h-x2x+h-x=limh0 -2sin x+h+x2 sinx+h-x22×x+h+xx+h-x2=limh0sinx+h-x2x+h-x2limh0 -sinx+h+x2x+h+x                       =1×-sinx2x                     limh0sinx+h-x2x+h-x2=1 =-sinx2x

(iii) Let f(x) = tanxThus, we have:(x+h)=tanx+hddx(f(x))=lim h0 f(x+h)-f(x) h=lim h0 tanx+h-tanx h=lim h0 sin x+h-xh cosx+h cos x           tan A-tan B= sin(A-B)cos A cos B      =lim h0      sin x+h-xx+h-x cosx+h cos x  =lim h0 sin x+h-xx+h-xx+h-xcosx+h cos x =lim h0sin x+h-xx+h-x.lim h0 1x+h+xcosx+hcosx           lim h0sinx+h-xx+h-x=1=1×12xcosxcosx=12xsec2x


(iv) Let f(x)= tan x2 Thus, we have:f(x+h) = tan (x+h)2ddxf(x)=limh0f(x+h)-f(x)h=limh0 tan (x+h)2-tanx2h=limh0sinx+h2-x2h cos x+h2cosx2              tan A-tan B=sin (A-B)cos A cos B=limh0 sin(x2+h2+2hx-x2)hcosx+h2cosx2=limh0sin(hh+2x)hh+2x cosx+h2cosx2×h+2x=limh0 sin(hh+2x)(hh+2x)limh0 h+2xcos(x+h)2cos x2         As limh0 sin(hh+2x)(hh+2x)=1=1×2xcos2 x2=2x sec2x2  



Page No 30.3:

Question 1:

Find the derivative of f (x) = 3x at x = 2

Answer:

We have:

f'(2) =lim                  h0f(2+h)-f(2)h        =limh03(2+h)-3(2)h        =limh06+3h-6h       =limh03hh       =3

Page No 30.3:

Question 2:

Find the derivative of f (x) = x2 − 2 at x = 10

Answer:

We have:
f'(x)=limh0f(10+h)-f(10)h       =limh0(10+h)2-2-(102-2)h      =limh0100+h2+20h-2-100+2h      =limh0h2+20hh      =limh0h(h+20)h     =limh0h+20    =0+20    =20

Page No 30.3:

Question 3:

Find the derivative of f (x) = 99x at x = 100

Answer:

We have:
f'(100)=limh0f(100+h)-f(100)h            =limh099(100+h)-99(100)h           =limh09900+99h-9900h          =limh099hh          =limh099           =99

Page No 30.3:

Question 4:

Find the derivative of f (x) x at x = 1

Answer:

We have:
f'(x)=limh0f(1+h)-f(1)h     =limh01+h-1h    =limh01    =1

Page No 30.3:

Question 5:

Find the derivative of f (x) = cos x at x = 0

Answer:

We have:
f'(x)=limh0f(0+h)-f(0)h=limh0f(h)-f(0)h=limh0cosh-cos0h=limh0cosh-1h=lim     h0 -2 sin2 h2h=limh0 -2 sin2 h2h24 ×h4==limh0 -1 ×h2=0

Page No 30.3:

Question 6:

Find the derivative of f (x) = tan x at x = 0

Answer:

We have:
f'(x)=limh0f(0+h)-f(0)h     =limh0f(h)-f(0)h    =limh0tanh-tan0h   =limh0tanhh   = 1

Page No 30.3:

Question 7:

Find the derivatives of the following functions at the indicated points:

(i) sin x at x = π2

(ii) x at x = 1

(iii) 2 cos x at x = π2

(iv) sin 2x at x = π2

Answer:

i We have:f'π2=limh0fπ2+h-fπ2h        =limh0sinπ2+h-sinπ2h      =limh0cos h-1h    =lim     h0 -2 sin2 h2h   =limh0 -2 sin2 h2h24 ×h4  =limh0-1 ×h2  =0
iiWe have:f'(x)=limh0f(1+h)-f(1)h      =limh01+h-1h    =limh01     =1

iii We have:f'π2=limh0fπ2+h-fπ2h          =limh02cosπ2+h-cosπ2h          =limh0-2sin h-0h          =-2limh0sinhh          =-2(1)          =-2
iv We have:f'π2=limh0fπ2+h-fπ2h          =limh0sin2π2+h-sin2π2h         =limh0sin(π+2h)-0h         =limh0-sin2hh×22             =limh0 -sin 2h2h×2         =-2



Page No 30.33:

Question 1:

x4 − 2 sin x + 3 cos x

Answer:

ddxx4-2 sin x+ 3 cos x=ddxx4-2ddx sin x+ 3ddxcos x=4x4-2 cos x - 3 sin x

Page No 30.33:

Question 2:

3x + x3 + 33

Answer:

ddx3x+x3+33=ddx3x+ddxx3+ddx33=3x log 3+3x2 + 0=3x log 3+3x2

Page No 30.33:

Question 3:

x33-2x+5x2

Answer:

ddxx33-2x+5x2=13ddxx3-2ddxx12+5ddxx-2=133x2-2.12.x-12+5-2x-3=x2-x-12-10x-3

Page No 30.33:

Question 4:

ex log a + ea long x + ea log a

Answer:

ddxex log a+ea log x+ea log a=ddxex log a+ddxea log x+ddxea log a=ddxelog ax+ddxelog xa+ddxelog aa=ddxax+ddxxa+ddxaa=ax log a + axa-1+0 =ax log a + axa-1

Page No 30.33:

Question 5:

(2x2 + 1) (3x + 2)

Answer:

ddx2x2+13x+2=ddx6x3+4x2+3x+2=6ddxx3+4ddxx2+3ddxx+ddx2=63x2+42x+31+0=18x2+8x+3

Page No 30.33:

Question 6:

 log3 x + 3 loge x + 2 tan x

Answer:

ddxlog3 x+ 3 loge x+2 tan x=ddxlog xlog 3  +3ddx loge x+2ddxtan x=1log 3.1x+3.1x+2 sec2 x=1x log 3+3x+2 sec2x



Page No 30.34:

Question 7:

x+1xx+1x

Answer:

ddxx+1xx+1x=ddxx+x-1x12+x-12=ddxx32+x12+x-12+x-32=ddxx32+ddxx12+ddxx-12+ddxx-32=32x12+12x-12-12x-32-32x-52

Page No 30.34:

Question 8:

x+1x3

Answer:

ddxx+1x3=ddxx3+3x21x+3x1x2+1x3=ddxx32+3ddxx12+3ddxx-12+ddxx-32=32x32-1+3.12x12-1+3-12x-12-1+-32x-32-1=32x12+32x-12-32x-32-32x-52 

Page No 30.34:

Question 9:

2x2+3x+4x

Answer:

ddx2x2+3x+4x=ddx2x2x+ddx3xx+ddx4x=2ddxx+3ddx1+4ddxx-1=21+30+4-1x-2=2-4x2

Page No 30.34:

Question 10:

(x3+1)(x-2)x2

Answer:

ddxx3+1x-2x2=ddxx4-2x3+x-2x2=ddxx4x2-2ddxx3x2+ddxxx2-ddx2x2=ddxx2-2ddxx+ddxx-1-2ddxx-2=2x-2-1x2-2-2x-3=2x-2-1x2+4x3

Page No 30.34:

Question 11:

a cos x+b sin x+csin x

Answer:

ddxa cos x+b sin x+csinx=ddxa cos xsin x+ddxb sin xsin x+ddxcsin x=addxcot x+ddxb+cddxcosec x=-a cosec2x+0-c cosec x cot x=-a cosec2x-c cosec x cot x

Page No 30.34:

Question 12:

2 sec x + 3 cot x − 4 tan x

Answer:

ddx2 sec x+ 3 cot x-4 tan x=2ddxsec x+3ddxcot x-4ddxtan x=2 sec x tan x-3 cosec2x-4 sec2x

Page No 30.34:

Question 13:

a0 xn + a1 xn−1 + a2 xn2 + ... + an1 x + an.

Answer:

ddxa0 xn+ddxa1 xn-1+ddxa2 xn-2+...+an-1 x+ an=a0ddxxn+a1ddxxn-1+a2ddxxn-2+...+an-1ddxx+ddxan=na0 xn-1+n-1 a1 xn-2+n-2a2 xn-3+....+an-11+0=na0 xn-1+n-1 a1 xn-2+n-2a2 xn-3+....+an-1

Page No 30.34:

Question 14:

1sin x+2x+3+4logx 3

Answer:

ddx1sin x+2x+3+4logx3=ddxcosec x+2x. 23+4log 3log x=ddxcosec x+23ddx2x+4log 3ddxlog x=-cosec x cot x+23.2x.log 2+4log 3.1x=-cosec x cot x+2x+3.log 2+4xlog 3

Page No 30.34:

Question 15:

(x+5)(2x2-1)x

Answer:

ddxx+52x2-1x=ddx2x3+10x2-x-5x=ddx2x3x+ddx10x2x-ddxxx-ddx5x=2ddxx2+10ddxx-ddx1-5ddxx-1=22x+101-0-5-1x-2=4x+10+5x2

Page No 30.34:

Question 16:

log1x+5xa-3ax+x23+6 x-34

Answer:

ddxlog 1x+5xa-3ax+x23+6 x-34=ddxlog x-12+5ddxxa-3ddxax+ddxx23+6ddxx-34=ddx-12log x+5ddxxa-3ddxax+ddxx23+6ddxx-34=-12.1x+5axa-1-3ax log a+23x-13+6-34x-74=-12x+5axa-1-3ax log a+23x-13-92x-74

Page No 30.34:

Question 17:

cos (x + a)

Answer:

ddxcos x+a=ddxcos x cos a - sin x sin a=cos addxcos x-sin a ddxsin x=-cos a sin x-sin a cos x=-sin x cos a + cos x sin a=-sinx+a

Page No 30.34:

Question 18:

cos (x-2)sin x

Answer:

ddxcosx-2sin x=ddxcos x cos 2+sin x sin 2sin x=ddxcos x cos 2sin x+ddxsin x sin 2sin x=cos 2ddxcot x+sin 2ddx1=cos 2 -cosec2x+sin 2 0=-cosec2x cos 2

Page No 30.34:

Question 19:

If y=sinx2+cosx22, find dydx at x=π6.

Answer:

dydx=ddxsin x2+cos x22       =ddxsin2 x2+cos2x2+2 sin x2cos x2      =ddx1+sinx      =ddx1+ddxsin x      =0+cos x      =cos xdydx at x = π6 = cos π6 = 32

Page No 30.34:

Question 20:

If y=2-3 cos xsin x, find dydx at x=π4

Answer:

dydx=ddx2-3 cos xsin x      =ddx2sin x-ddx3 cos xsin x      =2ddxcosec x-3ddxcot x      =-2 cosec x cot x+3 cosec2xdydx at x=π4=-2 cosec π4 cot π4+3 cosec2π4                      =-221+322                      =-22+6                      =6-22

Page No 30.34:

Question 21:

Find the slope of the tangent to the curve f (x) = 2x6 + x4 − 1 at x = 1.

Answer:

Slope of the tangent = f'(x)                                   =ddx2x6+x4-1                                   =2ddxx6+ddxx4-ddx1                                   =12x5+4x3 Slope of the tangent at x=1:1215+413=12+4=16

Page No 30.34:

Question 22:

If y=xa+ax, prove that 2xydydx=xa-ax

Answer:

y=xa+ax=1ax12+ax-12dydx=1a12x-12+a-12x-32LHS = 2xy dydx        = 2x 1ax12+ax-121a12x-12+a-12x-32       = 2x12a-12x+12x-a2x2        = 2x12a-a2x2        = xa-ax        = RHS Hence, proved.

Page No 30.34:

Question 23:

Find the rate at which the function f (x) = x4 − 2x3 + 3x2 + x + 5 changes with respect to x.

Answer:

Rate = f'(x)        =ddxx4-2x3+3x2+x+5        =ddxx4-2ddxx3+3ddxx2+ddxx+ddx5        =4x3-23x2+32x+1+0        =4x3-6x2+6x+1

Page No 30.34:

Question 24:

If y=2x93-57x7+6x3-x, find dydx at x=1.

Answer:

dydx=ddx2x93-57x7+6x3-x      =23ddxx9-57ddxx7+6ddxx3-ddxx      =239x8-577x6+63x2-1      =6x8-5x6+18x2-1dydx at x = 1: 618-516+1812-1=6-5+18-1=18

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Question 25:

If for f (x) = λ x2 + μ x + 12, f' (4) = 15 and f' (2) = 11, then find λ and μ.

Answer:

f'x=λddxx2+μddxx+ddx12f'x=2λx+μ       1Given: f'4=152λ4+μ=15     From 18λ+μ=15           2Also, given: f'2=112λ2+μ=11      From 14λ+μ=11                3Subtracting equation (3) from equation (2):4λ=4λ=1Substituting this in equation (3):41+μ=11μ=7∴ λ=1 and μ=7

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Question 26:

For the function f(x)=x100100+x9999+...+x22+x+1. Prove that f' (1) = 100 f' (0).

Answer:

f'x =ddxx100100+x9999+...+x22+x+1        =1100100 x99+19999 x98+...+122x+1+0        =x99+x98+...+x+1f'1=199+198+...+1+1        =99+1        =100f'0=0+0+...+0+1        =1RHS=100 f'0        =1001        =100        =f'1        =LHS f'1=100 f'0



Page No 30.39:

Question 1:

x3 sin x

Answer:

Let u=x3; v= sin xThen, u'= 3x2; v'=cos xUsing the product rule:ddxuv=uv'+vu'ddxx3 sin x=x3 cos x+ sin x 3x2                       =x2 x cos x+ 3 sin x

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Question 2:

x3 ex

Answer:

Let u= x3; v=exThen, u'=3x2; v'=exUsing the product rule:ddxuv=uv'+vu'ddxx3 ex=x3 ex + ex 3x2                   =x2ex x+3

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Question 3:

x2 ex log x

Answer:

Let u=x2; v=ex; w=log xThen, u'=2x; v'=ex, w= 1xUsing the product rule:ddxuvw=u'vw++uv'w+uvw'ddxx2ex log x=2x ex log x+x2 ex log x+x2 ex 1x                            =2x ex log x+x2 ex log x+x ex                            =x ex 2 log x+ x log x+1

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Question 4:

xn tan x

Answer:

Let u= xn; v=tan xThen, u'=nxn-1; v'=sec2xUsing the product rule:ddxuv=uv'+vu'ddxxn tan x=xn sec2x+tan xnxn-1                        =xn-1 x sec2x + n tan x

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Question 5:

xn loga x

Answer:

Let u=xn; v=loga x=log xlog aThen, u'=n xn-1; v'=1x log aUsing the product rule:ddxuv=uv'+vu'ddxxn loga x=xn . 1x log a+loga x n xn-1                          =xn-11 log a+loga x n xn-1                          =xn-1 1log a+n loga x

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Question 6:

(x3 + x2 + 1) sin x

Answer:

Let u=x3+x2+1; v= sin xThen,u'= 3x2+2x; v'=cos xBy product rule,ddxuv=uv'+vu'ddxx3+x2+1 sin x=x3+x2+1 cos x+3x2+2x sin x 

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Question 7:

sin x cos x

Answer:

Let u= sin x; v= cos xThen, u'= cos x; v'= - sin xUsing the product rule:ddxuv=uv'+vu'ddxsin x cos x=sin x -sin x+cos x . cos x                             = -sin2x+cos2x                             = cos 2x

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Question 8:

2x cot xx

Answer:

2x cot xx=2x cot x x-12Let u=2x; v=cot x; w=x-12Then, u'=2x log 2; v'=-cosec2 x; w'=-12x-32Using the product rule:ddxuvw=u'vw+uv'w+uvw'ddx2x cot x x-12 =2x log 2.cot x.x-12+2x-cosec2 x x-12+2x cot x-12x-32                                     =2x log 2.cot x.1x+2x-cosec2 x1x+2x cot x-12xx                                      =2xxlog 2 . cot x-cosec2x-cot x2x

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Question 9:

x2 sin x log x

Answer:

Let u=x2; v=sin x; w=log xThen, u'=2x; v'=cos x; w'=1xUsing the product rule:ddxuvw=u'vw+uv'w+uvw'ddxx2 sin x log x=2x sin x log x+x2 cos x log x+x2 sin x.1x                                 =2x sin x log x+x2 cos x log x+x sin x

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Question 10:

x5 ex + x6 log x

Answer:

ddxx5 ex + x6 log x=x5 ddxex+ex ddxx5+x6ddxlog x+ log x ddxx6=x5 ex +ex 5x4 +x6.1x+log x6x5=x5 ex +ex 5x4 +x5+log x6x5=x4 xex + 5ex +x+6x log x

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Question 11:

(x sin x + cos x) (x cos x − sin x)

Answer:

u= x sin x+cos x; v= x cos x-  sin xu'=x cos x+ sin x- sin x = x cos x ; v'=-  x sin x+cos x- cos x=-  x sin x             Using the product rule:ddxuv=uv'+vu'ddxx sin x+cos xx cos x-  sin x= x sin x+cos x-  x sin x + x cos x-  sin x x cos x                                                                   =-  x2 sin2 x-  x cos x sin x +x2 cos2 x-  x cos x sin x                                                                    =x2 cos2x -  sin2x-  x2 sin x cos x                                                                   =x2 cos 2x-  xsin 2x                                                                   = x x cos 2x- sin 2x

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Question 12:

(x sin x + cos x ) (ex + x2 log x)

Answer:

Let u=x sin x + cos x; v = ex + x2 log x Then, u' = xddxsin x+sinx  ddxx-sin x                  = x cos x +sinx - sinx                 = x cos x v' = ex +x2 ddxlog x + log x ddxx2       = ex +x + 2x log x                          Using the product rule: ddxuv= u v ' + v u'ddxx sinx + cos x ex + x2 cos x=x sin x + cos x ex +x + 2x log x   + ex + x2 log x  x cos x

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Question 13:

(1 − 2 tan x) (5 + 4 sin x)

Answer:

Let u=1-2 tan x; v = 5 + 4 sin x Then, u' = -2 sec2x; v' = 4 cos xUsing the product rule:ddxuv=u v' + v u'ddx1-2 tan x 5 + 4 sin x =1-2 tan x 4 cos x+ 5 + 4 sin x -2 sec2x= 4 cos x -8 ×sinxcosxcosx-10sec2x-8× sinxcos2x=4 cos x - 8 sin x - 10sec2 x-8 sec x tan x =4cos x- 2 sin x-52 sec2 x-2 sec x tan x

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Question 14:

(1 +x2) cos x

Answer:

Let u=1+x2; v= cos xThen ,u'=2x; v' =-sinxUsing the product rule:ddxuv=uv'+vu'ddx1+x2cos x=1+x2-sinx+ cos x2x                                   =-sinx - x2 sinx + 2x cos x 

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Question 15:

sin2 x

Answer:

ddxsin2 x=2 sin x ddxsin x    (Using the chain rule)= 2 sin x cos x= sin 2x

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Question 16:

logx2 x

Answer:

logx2 x=log xlog x2      (by change of base property)           =log x 2 log x        log x2=2 log x           =12Now ddxlogx2 x =ddx12                               =0           12 is a constant

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Question 17:

ex log x tan x

Answer:

Let u=ex; v= log x; w=tan xThen , u' = ex; v'= 1x×12x=12x; w' = sec2 xUsing the product rule:ddxuvw=u'vw + uv'w + uvw'                 =ex  log xtan x+ex×12xtan x+exlog x sec2 x                 =ex log x12.tan x+tanx2x+log x12.sec2 x                 =ex 12 log x. tan x+tanx2x+ 12 log x. sec2 x                  =ex2log x.tan x+tan xx+log x.sec2x

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Question 18:

x3 ex cos x

Answer:

Let u=x3; v=ex; w=cos xThen ,u'=3x2; v'=ex; w'=-sin xUsing the product rule:ddxuvw=u'vw+uv'w+uvw'ddxx3 ex cos x=3x2 ex cos x+x3 ex cos x+x3 ex -sin x                              =x2 ex 3 cos x+x cos x-x sin x

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Question 19:

x2 cosπ4sin x

Answer:

x2 cos π4sin x=x2 cos π4 cosec xLet u=x2; v=cos π4; w= cosec xThen, u'=2x; v'=0; w'=-cosec x cot xUsing the product rule:ddxuvw=u'vw+uv'w+uvw'ddxx2 cos π4 cosec x=2x cos π4cosec x+ x2 .0.cosec x+x2 cos π4-cosec x cot x                                           = cos π42x cosec x -x2 cosec x cot x                                           = cos π42xsin x -x2  cot xsin x

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Question 20:

x4 (5 sin x − 3 cos x)

Answer:

Let u=x4; v=5 sin x - 3 cos xThen, u'=4x3; v'= 5 cos x - 3 (-sinx) = 5 cos x + 3 sin x According to the product rule:ddxuv=u v'+v u'ddxx45 sin x - 3 cos x=x4  5 cos x + 3 sin x +5 sin x - 3 cos x 4x3                                              =x35x cos x + 3 x sin x + 20 sin x - 12 cos x                                              =x33x+20 sin x +5x - 12 cos x                                              =3 x4 sinx +20x3sin x+5x cos x -12 cos x 

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Question 21:

(2x2 − 3) sin x

Answer:

Let u=2x2-3; v=sin xThen,u'=4x; v'= cos xUsing the product rule:ddxuv=uv'+vu'ddx2x2-3 sin x =2x2-3 cos x+ 4x sin x 

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Question 22:

x5 (3 − 6x−9)

Answer:

Let u=x5; v=3-6x-9Then, u'=5x4; v'=54x-10Using the product rule:ddxuv=uv'+vu'ddxx53-6x-9=x554x-10+3-6x-95x4                               =54 x-5+15 x4-30x-5                               =15 x4+24x-5

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Question 23:

x4 (3 − 4x−5)

Answer:

Let u=x-4; v=3-4x-5Then, u'=-4x-5; v'=20 x-6Using the product rule:ddxuv=uv'+vu'ddxx-4 3-4x-5=x-4 20 x-6+3-4x-5-4x-5                                  =20x-10-12x-5+16x-10                                  =-12x-5 +36x-10

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Question 24:

x−3 (5 + 3x)

Answer:

Let u=x-3; v=5+3xThen, u=-3x-4; v'=3Using the product rule:ddxuv=uv'+vu'ddxx3 5+3x=x-3.3+5+3x -3x-4                            =3x-3 - 15 x-4 - 9 x-3                            =-15 x-4-6 x-3

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Question 25:

Differentiate in two ways, using product rule and otherwise, the function (1 + 2 tan x) (5 + 4 cos x). Verify that the answers are the same.

Answer:

Product rule (1st method):Let u=1+2 tan x; v=5+4 cos xThen, u'=2sec2x; v'=-4 sin xUsing the product rule:ddxuv=uv'+vu'ddx1+2 tan x5+4 cos x=1+2 tan x-4 sin x+5+4 cos x2sec2x                                                   =-4 sin x-8 tan x sin x+10 sec2x+8 sec x                                                   =-4 sin x+10 sec2x+8cosx- 8sin2xcos x                                                   = -4 sin x+10 sec2x+81-sin2xcos x                                                   =-4 sin x+10 sec2x+8cos2xcos x                                                   =-4 sin x+10 sec2x+8 cos x2nd method:1+2 tan x5+4 cos x=5+4 cos x+10 tan x+ 8 sin xNow, we have:ddx1+2 tan x5+4 cos x=ddx5+4 cos x+10 tan x+ 8 sin x                                                    =-4 sin x+10 sec2x+8 cos xUsing both the methods, we get the same answer.

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Question 26:

Differentiate each of the following functions by the product rule and the other method and verify that answer from both the methods is the same.
(i) (3x2 + 2)2
(ii) (x + 2) (x + 3)
(iii) (3 sec x − 4 cosec x) (−2 sin x + 5 cos x)

Answer:

i Product rule (1st method):Let u=3x2+2; v=3x2+2Then, u'=6x; v'=6xUsing the product rule:ddxuv=uv'+vu'ddx3x2+23x2+2=3x2+26x+3x2+26x                                      =18x3+12x+18x3+12x                                      =36x3+24x  2nd method:ddx3x2+22=ddx9x4+12x2+4                          =36x3+24xUsing both the methods, we get the same answer.

ii Product rule (1st method):Let u=x+2; v=x+3Then, u'=1; v'=1Using the product rule:ddxuv=uv'+vu'ddxx+2x+3=x+21+x+31                               =x+2+x+3                               =2x+52nd method:ddxx+2x+3=ddxx2+5x+6                                =2x+5Using both the methods, we get the same answer.

iii Product rule (1st method):Let u=3 sec x-4 cosec x; v=-2 sin x+ 5 cos xThen, u'=3 sec x tan x+4 cosec x cot x; v'=- 2 cos x-5 sin xUsing the product rule:ddxuv=uv'+vu'ddx3 sec x-4 cosec x-2 sin x+ 5 cos x=3 sec x-4 cosec x- 2 cos x-5 sin x+-2 sin x+ 5 cos x3 sec x tan x+4 cosec x cot x                                                                                 =-6+15 tan x+8  cot x+20 -6 tan2x-8 cot x-15 tan x+ 20  cot2x                                                                                 =-6+20 -6sec2x -1+ 20  cosec2x-1                                                                                 =-6+20 -6sec2x +6+ 20 cosec2x-20                                                                                 =-6 sec2x+20 cosec2x2nd method:ddx3 sec x-4 cosec x-2 sin x+ 5 cos x=ddx-6 sec x sin x +15 sec x cos x+8 cosec x sin x -20 cosec x cos x                                                                                 =ddx-6 sin xcos x +15cos xcos x+8 sin xsin x -20 cos xsin x                                                                                 =ddx-6 tan x+15 +8 -20 cot x                                                                                 =ddx-6tan x-20 cot x+23                                                                                 =-6 sec2x+20 cosec2xUsing both the methods, we get the same answer.

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Question 27:

(ax + b) (a + d)2

Answer:

(ax+b)(a+d)2Let u =ax+b, v= a+d2Then, u'=a, v'= 0Using the product rule:ddxuv=u v' + v u'ddx(ax+b)(a+d)2=(ax+b)×0+ a+d2×a ddx(ax+b)(a+d)2=aa+d2

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Question 28:

(ax + b)n (cx + d)n

Answer:

ax+bn cx+dnLet u =ax+bn, v=cx+dnThen, u'=naax+bn-1, v'=nccx+dn-1Using the product rule:ddxuv=uv'+u'vddxax+bn cx+dn=ax+bn×nccx+dn-1+naax+bn-1×cx+dn                                          =nax+bn-1cx+dn-1acx+cb+acx+ad                                          =nax+bn-1cx+dn-12acx+cb+ad



Page No 30.44:

Question 1:

x2+1x+1

Answer:

Let u=x2+1; v=x+1Then, u'=2x; v'=1Using the quotient rule:ddxuv=vu'-uv'v2ddxx2+1x+1=x+12x-x2+11(x+1)2                       =2x2+2x-x2-1(x+1)2                       =x2+2x-1(x+1)2

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Question 2:

2x-1x2+1

Answer:

Let u=2x-1; v=x2+1;Then, u'=2; v'=2xUsing the quotient rule:ddxuv=vu'-uv'v2ddx2x-1x2+1=x2+12-2x-12x(x2+1)2                        =2x2+2-4x2+2x(x2+1)2                        =-2x2+2x+2(x2+1)2                        =21+x-x2(x2+1)2

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Question 3:

x+ex1+log x

Answer:

Let u=x+ex; v=1+logxThen, u'=1+ex; v'=1xUsing the quotient rule:ddxuv=vu'-uv'v2ddxx+ex1+logx=1+log x1+ex-x+ex1x(1+log x)2                           =x+xex+x log x+x log x ex-x-exx(1+logx)2                           =x log x+ x log x ex-ex+x exx(1+logx)2                           =x log x 1+ex-ex1-xx(1+logx)2

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Question 4:

ex-tan xcot x-xn

Answer:

Let u=ex-tan x; v=cot x-xnThen, u'=ex-sec2x; v'=-cosec2x-nxn-1Using the quotient rule:ddxuv=vu'-uv'v2ddxex-tan xcot x-xn=cot x-xnex-sec2x-ex-tan x-cosec2x-nxn-1cot x-xn2                              =cot x-xnex-sec2x+ex-tan xcosec2x+nxn-1cot x-xn2

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Question 5:

ax2+bx+cpx2+qx+r

Answer:

Let u=ax2+bx+c; v=px2+qx+rThen, u'=2ax+b; v'=2px+qUsing the quotient rule:ddxuv=vu'-uv'v2ddxax2+bx+cpx2+qx+r=px2+qx+r2ax+b-ax2+bx+c2px+qpx2+qx+r2                                  =2apx3+2aqx2+2arx+bpx2+bqx+br-2apx3-2bpx2-2pcx-aqx2-bqx-cqpx2+qx+r2                                  =aq-bpx2+2ar-xpx+br-cqpx2+qx+r2

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Question 6:

x1+tan x

Answer:

Let u=x; v=1+tan xThen, u'=1; v'= sec2xUsing the quotient rule:ddxuv=vu'-uv'v2                =1+tan x×1-xsec2x1+tanx2                =1+tan x-xsec2x1+tanx2

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Question 7:

1ax2+bx+c

Answer:

ddx1ax2+bx+c=ddxax2+bx+c-1=-1ax2+bx+c-2ddxax2+bx+c    (Using the chain rule)=-1ax2+bx+c-22ax+b=-2ax+bax2+bx+c2

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Question 8:

ex1+x2

Answer:

Let u=ex; v=1+x2Then, u'=ex; v'=2xUsing the chain rule:ddxuv=vu'-uv'v2ddxex1+x2=1+x2ex-ex2x1+x22                       =ex+x2ex-2xex1+x22                       =ex1+x2-2x1+x22                       =ex1-x21+x22

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Question 9:

ex+sin x1+log x

Answer:

Let u=ex+sin x; v=1+ log xThen, u'=ex+cos x; v'= 1xUsing the quotient rule:ddxuv=vu'-uv'v2ddxex+sin x1+ log x=1+log xex+cos x-ex+sin x 1x1+ log x2                              =x1+log xex+cos x-ex+sin xx1+ log x2

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Question 10:

x tan xsec x+tan x

Answer:

Let u=x tan x; v=sec x+tan xThen, u'=x sec2x+tan x; v'=sec x tan x + sec2xUsing the quotient rule:ddxuv=vu'-uv'v2ddxxtan x sec x +tan x =sec x+tan xx sec2x+tan x-x tan xsec x tan x + sec2xsec x+tan x2                                     =x sec3x+x sec2xtanx+secx tanx+tan2x-x sec x tan2x-x tanx sec2xsec x+tan x2                                     =sec x+tan xx sec2x+tan x-x tan x secxsec x+tan xsec x+tan x2                                     =x sec2x+tan x-x tanx secxsec x+ tan x                                       =x sec xsec x -tan x+tan xsec x + tan x

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Question 11:

x sin x1+cos x

Answer:

Let u=x sinx; v=1+ cos xThen, u' = x cos x + sinx; v' = -sin xUsing the quotient rule:ddxuv=vu'-uv'v2ddxx sin x1+cos x=1+ cos xx cos x + sinx-x sinx-sin x1+ cos x2                             =1+ cos xx cos x + sinx+x sin2 x1+ cos x2                             =1+ cos xx cos x + sinx+x 1-cos2x1+ cos x2                             =1+ cos xx cos x + sinx+x1+cos x1-cos x1+ cos x2                             =1+ cos xx cos x + sinx+x- xcosx1+ cos x2                             =1+ cos xx+sin x1+ cos x2

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Question 12:

2x cot xx

Answer:

Let u=2xcot x; v=xThen, u'=-2xcosec2x+2xlog 2 cot x; v'=12xddxuv=vu'-uv'v2ddx2xcot xx=x-2xcosec2x+2xlog 2 cot x-2xcot x12xx2                          =x-2xcosec2x+2xlog 2 cot x-2x-1cot xxx                          =2x-xcosec2x+x cot x log 2 -12cot xxx                          =2x-xcosec2x+x cot x log 2 -12cot xx32

Page No 30.44:

Question 13:

sin x-x cos xx sin x+cos x

Answer:

Let u=sin x-x cos x; v=x sinx + cos xThen, u'=cosx+x sin x-cosx; v'=x cos x+sin x-sinx   =x sin x                           =x cos xUsing the quotient rule:ddxuv=vu'-uv'v2ddxsin x-x cos xx sinx + cos x=x sinx + cos xx sinx -sin x-x cos xx cos xx sinx + cos x2                                       =x2sin2x+x cosx sinx -x cos x sin x+x2cos2xx sinx + cos x2                                       =x2sin2x+cos2xx sinx + cos x2                                       =x2x sinx + cos x2

Page No 30.44:

Question 14:

x2-x+1x2+x+1

Answer:

Let u=x2-x+1; v= x2+x+1Then, u'=2x-1; v'=2x+1By quotient rule,ddxuv=vu'-uv'v2ddxx2-x+1x2+x+1=x2+x+12x-1-x2-x+12x+1x2+x+12                              =2x3+2x2+2x-x2-x-1-2x3+2x2-2x-x2+x-1x2+x+12                             =2x2-2x2+x+12                             =2x2-1x2+x+12

Page No 30.44:

Question 15:

a+xa-x

Answer:

Let u=a+x; v=a-xThen,u'=12x; v'=-12xUsing the quotient rule:ddxuv=vu'-uv'v2ddxa+xa-x=a-x12x-a+x-12xa-x2                              =a-x+a+x2xa-x2                              =2a2xa-x2                             =axa-x2

Page No 30.44:

Question 16:

a+sin x1+a sin x

Answer:

Let us use the quotient rule here.
We have:
u = a + sin x and v =1 + a sin x
u' = cos x and v'=a cos x

Using the quotient rule:ddxuv=vu'-uv'v2ddxa+sinx1+asinx=(1+asinx)(cosx)-(a+sinx)(acosx)(1+asin x)2                             =cosx+asinx cosx -a2cosx-a sinx cosx(1+asin x)2                             =cosx-a2cosx(1+asinx)2                            =(1-a2)cosx(1+a sinx)2

Page No 30.44:

Question 17:

10xsin x

Answer:

Let u=10x; v=sin xThen,u'=10xlog 10; v'= cos xUsing the quotient rule:ddxuv=vu'-uv'v2ddx10xsin x=sin x 10xlog 10-10x cos xsin2x                      =sin x 10xlog 10sin2x-10x cos xsin2x                      =10xlog 10 cosec x -10xcosec x cot x                      =10x cosec xlog 10 - cot x

Page No 30.44:

Question 18:

1+3x1-3x

Answer:

Let u=1+3x; v=1-3xThen,u'=3x log 3; v'=-3x log 3Using the quotient rule:ddxuv=vu'-uv'v2ddx1+3x1-3x=1-3x3x log 3-1+3x-3x log 31-3x2                       =3x log 3-32x log 3+3x log 3+32x log 31-3x2                       =2 . 3x log 31-3x2

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Question 19:

3xx+tan x

Answer:

Let u=3x; v=x+tan xThen, u'=3xlog 3; v'=1+sec2xBy quotient rule, we have:ddxuv=vu'-uv'v2ddx3xx+tan x=x+tan x3xlog 3-3x 1+sec2xx+tan x2                            =3xx+tan x log 3- 1+sec2xx+tan x2

Page No 30.44:

Question 20:

1+log x1-log x

Answer:

Let u=1+log x; v=1- log xThen, u'=1x; v'=-1xUsing the quotient rule:ddxuv=vu'-uv'v2ddx1+log x1- log x=1- log x1x-1+log x-1x1- log x2                             =1-log x+1+log xx1- log x2                             =2 x1- log x2

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Question 21:

4x+5 sin x3x+7 cos x

Answer:

Let u=4x + 5 sin x; v=3x + 7 cos xThen, u'=4+5 cos x; v'=3-7 sin xUsing the quotient rule:ddxuv=vu'-uv'v2ddx4x + 5 sin x3x + 7 cos x=3x + 7 cos x4+5 cos x-4x + 5 sin x3-7 sin x3x + 7 cos x2                                    =12x+15 x cos x+28 cos x+35 cos2x-12x +28 x sin x-15 sin x +35 sin2x3x + 7 cos x2                                    =15 x cos x+ 28 x sin x+28 cos x15 sinx +35sin2x+cos2x3x + 7 cos x2                                    =15 x cos x+ 28 x sin x+28 cos x15 sinx +353x + 7 cos x2

Page No 30.44:

Question 22:

x1+tan x

Answer:

Let u=x; v=1+tan xThen, u'=1; v'= sec2xUsing the quotient rule:ddxuv=vu'-uv'v2ddxx1+tan x=1+tan x1-xsec2x1+tan x2                            =1+tan x-xsec2x1+tan x2

Page No 30.44:

Question 23:

a+b sin xc+d cos x

Answer:

Let u=a+b sin x; v=c+d cos xThen, u'=b cos x; v'=-d sin xUsing the quotient rule:ddxuv=vu'-uv'v2ddxa+b sin xc+d cos x=c+d cos xb cos x-a+b sin x-d sin xc+d cos x2                                 =bc cos x+bd cos2x+ad sin x+bd sin2xc+d cos x2                                  =bc cos x+ad sin x+bd sin2x+cos2xc+d cos x2                                  =bc cos x+ad sin x+bd c+d cos x2

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Question 24:

px2+qx+rax+b

Answer:

Let u=px2+qx+r; v=ax+bThen, u'=2px+q; v'=aUsing the quotient rule:ddxuv=vu'-uv'v2ddxpx2+qx+rax+b=ax+b2px+q-px2+qx+raax+b2                                 =2ap x2+aq x+2bp x+bq-ap x2-aq x-arax+b2                                 =ap x2+2bp x+bq -arax+b2

Page No 30.44:

Question 25:

sec x-1sec x+1

Answer:

Let u=sec x-1; v=sec x+1Then,u'=sec x tan x; v'=sec x tan xUsing the quotient rule:ddxuv=vu'-uv'v2ddxsec x-1sec x+1=sec x+1sec x tan x-sec x-1sec x tan xsec x+12                             =sec2x tan x+ sec x tan x-sec2x tan x+sec x tan xsec x+12                             =2sec x tan xsec x+12

Page No 30.44:

Question 26:

x5-cos xsin x

Answer:

Let u=x5-cos x; v=sin xThen, u'= 5x4+sin x; v'=cos xUsing the quotient rule:ddxuv=vu'-uv'v2ddxx5-cos xsin x=sin x 5x4+sin x-x5-cos xcos xsin2x                               =-x5cos x + 5x4sin x+sin2x+cos2xsin2x                               =-x5cos x + 5x4sin x+1sin2x

Page No 30.44:

Question 27:

x+cos xtan x

Answer:

Let u=x+cos x; v=tan xThen, u'=1-sin x; v'= sec2xUsing the quotient rule:ddxuv=vu'-uv'v2ddxx+cos xtan x=tan x1-sin x-x+cos xsec2xtan2x

Page No 30.44:

Question 28:

xsinn x

Answer:

Let u=x; v=sinnxThen,u'=1; v'=n sinn-1x.cos xUsing the quotient rule:ddxuv=vu'-uv'v2ddxxsinnx=sinnx.1-x n sinn-1 x.cos xsinnx2                      =sinn-1xsinx-nx.cos xsin2n x                      =sinx-nx.cos xsin2n-n-1 x                      =sinx-nxcos xsinn+1 x

Page No 30.44:

Question 29:

ax+bpx2+qx+r

Answer:

Let u=ax+b; v=px2+qx+rThen,u'=a; v'=2px+qUsing the quotient rule:ddxuv=vu'-uv'v2ddxax+bpx2+qx+r=px2+qx+ra-ax+b2px+qpx2+qx+r2                                 =ap x2+aq x+ar-2ap x2-2bp x-aq x-bqpx2+qx+r2                                 =-apx2-2bp x+ar-bqpx2+qx+r2

Page No 30.44:

Question 30:

1ax2+bx+c

Answer:

Let u=1; v=ax2+bx+cThen,u'=0; v'=2ax+bUsing the quotient rule:ddxuv=vu'-uv'v2ddx1ax2+bx+c=ax2+bx+c0-12ax+bax2+bx+c2                                  =-2ax+bax2+bx+c2



Page No 30.46:

Question 1:

Mark the correct alternative in each of the following:

Let f(x) = x − [x], xR, then f'12 is

(a) 32                                  (b) 1                                  (c) 0                                  (d) −1

Answer:

Given: f(x) = x − [x], xR

Now,

For 0 ≤ x < 1, [x] = 0.

f(x) = x − 0 = x, ∀ x ∈ [0, 1)

Differentiating both sides with respect to x, we get

f '(x) = 1, ∀ x ∈ [0, 1)

 f'12=1

Hence, the correct answer is option (b).



Page No 30.47:

Question 2:

Mark the correct alternative in each of the following:

If fx=x-42x, then f '(1) is

(a) 54                                  (b) 45                                  (c) 1                                  (d) 0

Answer:


fx=x-42x      =12x-2x      =12x12-2x-12

Differentiating both sides with respect to x, we get

f'x=12×12x12-1-2×-12x-12-1                    fx=xnf'x=nxn-1f'x=14x-12+x-32f'1=14×1+1=54

Hence, the correct answer is option (a).

Page No 30.47:

Question 3:

Mark the correct alternative in each of the following:

If y=1+x1!+x22!+x33!+..., then dydx=

(a) y + 1                                 (b) y − 1                                  (c) y                                 (d) y2

Answer:

y=1+x1!+x22!+x33!+...

Differentiating both sides with respect to x, we get

dydx=ddx1+x1!+x22!+x33!+...       =ddx1+ddxx1!+ddxx22!+ddxx33!+ddxx44!+...       =ddx1+11!ddxx+12!ddxx2+13!ddxx3+14!ddxx4+...       =0+11!×1+12!×2x+13!×3x2+14!×4x3+...                                    y=xndydx=nxn-1
       =1+x1!+x22!+x33!+...                                                          nn!=1n-1!=y

dydx=y

Hence, the correct answer is option (c).

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Question 4:

Mark the correct alternative in each of the following:

If fx=1-x+x2-x3+...-x99+x100, then f'1 equals

(a) 150                                  (b) −50                                  (c) −150                                  (d) 50

Answer:

fx=1-x+x2-x3+...-x99+x100

Differentiating both sides with respect to x, we get

f'x=ddx1-x+x2-x3+...-x99+x100       =ddx1-ddxx+ddxx2-ddxx3+...-ddxx99+ddxx100       =0-1+2x-3x2+...-99x98+100x99       =-1+2x-3x2+...-99x98+100x99

Putting x = 1, we get

f'1=-1+2-3+...-99+100        =-1+2+-3+4+-5+6+...+-99+100        =1+1+1+ ... +1   50 terms        =50

Hence, the correct answer is option (d).

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Question 5:

Mark the correct alternative in each of the following:

If y=1+1x21-1x2, then dydx=

(a) -4xx2-12                                  (b) -4xx2-1                                  (c) 1-x24x                                  (d) 4xx2-1

Answer:


y=1+1x21-1x2  =x2+1x2-1

Differentiating both sides with respect to x, we get

dydx=x2-1×ddxx2+1-x2+1×ddxx2-1x2-12                Quotient rule       =x2-1×2x+0-x2+1×2x-0x2-12       =2x3-2x-2x3-2xx2-12       =-4xx2-12

Hence, the correct answer is option (a).

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Question 6:

Mark the correct alternative in each of the following:

If y=x+1x, then dydx at x = 1 is

(a) 1                                  (b) 12                                  (c) 12                                  (d) 0

Answer:


y=x+1x  =x12+x-12

Differentiating both sides with respect to x, we get

dydx=ddxx12+x-12       =ddxx12+ddxx-12       =12x12-1+-12x-12-1                               y=xndydx=nxn-1       =12x-12-12x-32

Putting x = 1, we get

dydxx=1=12×1-12×1=0

Thus, dydx at x = 1 is 0.

Hence, the correct answer is option (d).

Page No 30.47:

Question 7:

Mark the correct alternative in each of the following:

If fx=x100+x99+ ... +x+1, then f'1 is equal to

(a) 5050                                  (b) 5049                                  (c) 5051                                  (d) 50051

Answer:


fx=x100+x99+ ... +x+1

Differentiating both sides with respect to x, we get

f'x=ddxx100+x99+ ... +x+1       =ddxx100+ddxx99+ ... +ddxx2+ddxx+ddx1       =100x99+99x98+ ... +2x+1+0                                                       y=xndydx=nxn-1       =100x99+99x98+ ... +2x+1

Putting x = 1, we get

f'1=100+99+98+ ... +2+1        =100100+12                          Sn=nn+12        =50×101        =5050

Hence, the correct answer is option (a).

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Question 8:

Mark the correct alternative in each of the following:

If fx=1+x+x22+ ... +x100100, then f'1 is equal to

(a) 1100                                  (b) 100                                  (c) 50                                  (d) 0

Answer:


fx=1+x+x22+ ... +x100100

Differentiating both sides with respect to x, we get

f'x=ddx1+x+x22+ ... +x100100        =ddx1+ddxx+ddxx22+ ... +ddxx100100        =ddx1+ddxx+12ddxx2+ ... +1100ddxx100        =0+1+12×2x+ ... +1100×100x99                                 y=xndydx=nxn-1            =1+x+x2+ ... +x99

Putting x = 1, we get

f'1=1+1+1+ ... +1   100 terms        =100

Hence, the correct answer is option (b).

Page No 30.47:

Question 9:

Mark the correct alternative in each of the following:

If y=sinx+cosxsinx-cosx, then dydx at x = 0 is

(a) −2                                 (b) 0                                  (c) 12                                  (d) does not exist

Answer:

y=sinx+cosxsinx-cosx

Differentiating both sides with respect to x, we get

dydx=sinx-cosx×ddxsinx+cosx-sinx+cosx×ddxsinx-cosxsinx-cosx2                    Quotient rule      =sinx-cosx×ddxsinx+ddxcosx-sinx+cosx×ddxsinx-ddxcosxsinx-cosx2      =sinx-cosxcosx-sinx-sinx+cosxcosx+sinxsinx-cosx2      =-cos2x+sin2x-2cosx sinx-sin2x+cos2x+2sinx cosxsinx-cosx2
      =-1+2cosx sinx-1-2sinx cosxsinx-cosx2=-2sinx-cosx2

Putting x = 0, we get

dydxx=0=-2sin0-cos02=-20-12=-2

Thus, dydx at x = 0 is −2.

Hence, the correct answer is option (a).

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Question 10:

Mark the correct alternative in each of the following:

If y=sinx+9cosx, then dydx at x = 0 is

(a) cos 9                                  (b) sin 9                                  (c) 0                                  (d) 1

Answer:


y=sinx+9cosx

Differentiating both sides with respect to x, we get

dydx=cosx×ddxsinx+9-sinx+9×ddxcosxcos2x                Quotient rule       =cosx×cosx+9-sinx+9×-sinxcos2x       =cosx+9cosx+sinx+9sinxcos2x       =cosx+9-xcos2x       =cos9cos2x

Putting x = 0, we get

dydxx=0=cos9cos20=cos9                        (cos 0 = 1)

Thus, dydx at x = 0 is cos 9.

Hence, the correct answer is option (a).

Page No 30.47:

Question 11:

Mark the correct alternative in each of the following:

If fx=xn-anx-a, then f'a is

(a) 1                                  (b) 0                                  (c) 12                                  (d) does not exist

Answer:


Given: fx=xn-anx-a

Now, f(x) is not defined at x = a. Therefore, f(x) is not differentiable at x = a.

So, f'a does not exist.

Hence, the correct answer is option (d).

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Question 12:

Mark the correct alternative in each of the following:

If f(x) = x sinx, then f'π2= 

(a) 0                                  (b) 1                                  (c) −1                                  (d) 12

Answer:


f(x) = x sinx

Differentiating both sides with respect to x, we get

f'x=x×ddxsinx+sinx×ddxx                            Product rule       =x×cosx+sinx×1       =x cosx+sinx

Putting x=π2, we get

f'π2=π2×cosπ2+sinπ2            =π2×0+1            =1

Hence, the correct answer is option (b).



Page No 30.48:

Question 1:

If = 1 +x1!+x22!+x33!+......, then dydx = ______________________________.

Answer:

If y = 1 + x1!+x22!+x33!+......
dydx =ddx1+x1!+x22!+x33!+......          = ddx1+ddxx1!+ddxx22!+ddxx33!+ ....           = 0+11! ddxx+12! ddxx2+13! ddxx3+.....           = 0+11!×1+12!×2x+13!×3x2+......            = 1+x1!+13×2!3x2+14×3!4x3+.....i.e dydx= 1+x1!+12!x2+13!x3+......

Page No 30.48:

Question 2:

If y=1-x1!+x22!-x33!+x44!-x55!+...., then dydx= ________________________.

Answer:

If y=1-x1!+x22!-x33!+x44!-x55!+.......  
Then dydx=ddx1-x1!+x22!-x33!+x44!-....... 
            =ddx1-ddxx1!+ddxx22!-ddxx33!+...........            =0-11!ddxx+12!ddxx2-13!ddxx3+..........            =-11!×1+12!×2x-13!×3x2+14!×4x3-........            =-11!+x1!-12!x2+13!x3-...........i.e dydx=-1-x1!+x22!-x33!+x44!-........

Page No 30.48:

Question 3:

  Let f(x) = x – [x], x ∈ R. Then '13 =__________________.

Answer:

Given, f(x) = x – [x]
For 13 we know 13 = 0
i.e  f(x) = x – [x] = {x} = x      since x13
∴  f'(x) = 1  at x = 13
i.e  f'13 = 1

Page No 30.48:

Question 4:

If f(x) = xx, then f'(1) = _________________________.

Answer:

for f(x) = x + x
for x = 1, x = x
i.e  f(x) = x + x
i.e  f(x) = 2x
i.e  f'(x) = 2   at x = 1
i.e  f'(1) = 2

Page No 30.48:

Question 5:

If f(x) = x + x, then f'(–1) = _________________________.

Answer:

For f(x) = x + x given
at x = –1, x = – x
f(x) = x – x = 0
i.e  f'(x) = 0  at x = –1
i.e  f'(–1) = 0

Page No 30.48:

Question 6:

If f(x) = x x, then f'(–2) = _________________________.

Answer:

For f(x) = x x given
at x = –2
x = – x
∴ f(x) = x (– x)
i.e  f(x) = – x2
f'(x) = –2x at x = –2
i.e  f'(x) = (–2) (–2)
i.e  f'(x) = 4

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Question 7:

ddxx+1x2=______________.

Answer:

ddxx+1x2
=ddxx+1x+2x1x=ddxx+1x+2=ddxx+ddxx-1+ddx2=1-1x-1-1 +0i.e ddxx+1x2=1-1x2

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Question 8:

ddx(logx)=_______________, x0.

Answer:

ddxlogx
case (i) if x > 0
i.e x = x
⇒ ddxlog x=1x
case (ii) if x < 0
i.e x = – x
ddxlog-x=1-x×ddx-x=-1x×-1=1x

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Question 9:

If f(x) = mx + c,  f(0) = f'(0) = 1, then f(2) = ___________________.

Answer:

Given f(x) = mx + c
Since f(0) = 1
f(0) = m(0) + c
i.e 1 = m(0) + c
c = 1
i.e f(x) = mx + 1
f'(x) = (mx + 1)'
           =
(mx)' +(1)'
           = m
(x)' +(1)'
           = m ×
1 + 0
i.e f'(x) = m
also, given f'(0) = 1
f'(0) = m
i.e 1 = m
f(x) = x +
i.e f(2) = 2 + 1 = 3

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Question 10:

ddx1logxe=________________.

Answer:

ddx1logxe
Since logxe      =logeelogex=1logex=ddx11logex= ddxlogexi.eddx1logxe =1x

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Question 11:

ddx1log10x=______________.

Answer:

Check Answer
ddx1log10x
Since log10x=logexloge10i.e ddxloge10logex

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Question 12:

ddxlogxnxm=_____________.

Answer:

ddxlogxnxm
Since logxnxm  = logexmlogexn                        = m logexn logex By property log, log pq=q log p q log pi.e logxnxm      = mnddxlogxnxm=ddxmnddxlogxnxm=0

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Question 1:

Write the value of limxcf(x)-f(c)x-c.

Answer:

Using the definition of derivative, we have:limxcfx-fxx-c=f'c

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Question 2:

Write the value of limxax f (a)-a f (x)x-a.

Answer:

limxaxfa-afxx-a=limxaxfa-afx-xfx+xfxx-a=limxaxfa-xfx+xfx-afxx-a=limxa-xfx-fa+x-afxx-a=limxa-xlimxafx-fax-a+limxax-afxx-a=-a f'a+f(a)

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Question 3:

If x < 2, then write the value of ddx(x2-4x+4).

Answer:

Given: x<2∴ 2-x>0ddxx2-4x+4=ddx2-x2=ddx2-x   (∵ 2-x>0)=0-1=-1

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Question 4:

If π2 < x < π, then find ddx1+cos 2x2.

Answer:

1+cos 2x2=2 cos2x2=cos2x=-cos x    (∵ π2<x<π)ddx1+cos 2x2=ddx-cosx=--sinx=sinx



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Question 5:

Write the value of ddx x x .

Answer:

Case 1:x>0|x|=xThus, we have: ddxx|x|=ddxx.x=ddxx2=2x           1Case 2:x<0|x|=-xThus, we have: ddxx|x|=ddxx.-x=ddx-x2=-2x           2From (1) and (2), we have:ddxx|x|=2x, if x>0-2x, if x<0

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Question 6:

Write the value of ddx x+x x.

Answer:

Case 1:x>0x=xx+xx=x+xx=2x2ddxx+xx=ddx2x2=4x         1Case 2:x<0x=-xx+xx=x-xx=0ddxx+xx=ddx0=0            2From (1) and (2), we have:ddxx+xx=4x, if x>00, if x<0

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Question 7:

If f (x) = |x| + |x−1|, write the value of ddxf (x).

Answer:

fx=x+x-1Case 1: x<0 (∴ x-1<-1<0)x=-x; x-1=-x-1=-x+1fx=-x+-x+1=-2xf'x=-2Case 2: 0< x <1 (∴ x>0 and x-1<0)x=x; x-1=-x-1=1-xfx=x+1-x=1f'x=0Case 3: x>1 ∴ x>1>0 ⇒ x>0)x=x; x-1=x-1fx=x+x-1=2x-1f'x=2From case 1, case 2 and case 3, we have:f'x=-2, when x<0    0, when 0<x<12, when x>1

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Question 8:

Write the value of the derivative of f (x) = |x − 1| + |x − 3| at x = 2.

Answer:

Let x = 2We know: 2>1 and 2<3∴ x>1 and x<3x-1=x-1 and x-3=-x-3=-x+3fx=x-1+x-3=x-1-x+3=2f'x=0

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Question 9:

If f (x) = x2x, write ddxf (x).

Answer:

Case 1: x>0x=xfx=x2x=x2x=xf'x=1Case 2: x<0x=-xfx=x2x=x2-x=-xf'x=-1From case 1 and case 2, we have:f'x=1, if x>0-1, if x<0

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Question 10:

Write the value of ddxlog x.

Answer:

Case 1: x>0x=x   ...1ddxlog x=log x                  =1x                  =1x    (from (1))Case 2: x<0x=-x     ...2ddxlog x=log -x                   =1-x                   =1x  (from (2))From case (1) and case(2), ddxlog x=1x

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Question 11:

If f (1) = 1, f' (1) = 2, then write the value of limx1f (x)-1x-1.

Answer:

limx1 fx-1x-1=limx1 fx-1x-1×fx+1fx+1×x+1x+1=limx1 fx-1x+1x-1fx+1=limx1 fx-1x-1× limx1x+1fx+1=f'1×1+1f1+1=2×21+1=2

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Question 12:

Write the derivative of f (x) = 3 |2 + x| at x = −3.

Answer:

Let x = -3We know:-3<-2Thus, we have: x<-2It gives x+2<0. 2+x=x+2=-x+2=-x-2fx=3 2+x=3-x-2=-3x-6f'x=-3ddxx-ddx6=-3

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Question 13:

If |x| < 1 and y = 1 + x + x2 + x3 + ..., then write the value of dydx.

Answer:

The given series is a geometric series where a = 1 and r = x.
fx=1+x+x2+x3+...=11-xSum of the infinite series of a geometric series is a1-r.f'x=-1(1-x)2ddx(1-x)=-1(1-x)2(-1)=1(1-x)2

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Question 14:

If f (x) = logx2 x3, write the value of f' (x).

Answer:

f(x)=logx2x3      =logx3logx2   (Change of base property)      =3 logx2 log x      =32f'x=0 (Since 32 is a constant)



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