wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

(i) − x

(ii) (−x)−1

(iii) sin (x + 1)

(iv) cosx-π8

Open in App
Solution

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

ii -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

iii 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

iv 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

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Method of Substitution
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon