‌∴f′(0−)=−1‌ and ‌f′(0+)=1 ‌⇒f′(0−)≠f′(0+) So, it is not differentiable at x=0. (b) We have, f(x)‌=cos|x|−|x| ‌={
cos(−x)−(−x)
‌
x<0
cos(x)−x
‌
x≥0
. ‌={
cos‌x+x
‌
x<0
cos‌x−x
‌
x≥0
.
∴f′(0−)‌=1‌ and ‌f′(0+)=−1 ⇒f′(0−)≠f′(0+) So, it is not differentiable at x=0. (c) We have, ‌f(x)=sin‌|x|+|x|
‌={
−sin‌x−x
‌
x<0
sin‌x+x
‌
x≥0
. ‌∴‌‌f′(x)={
−cos‌x−1
‌
x<0
cos‌x+1
‌
x≥0
. ‌∴‌‌f′(0−)=−1−1=−2 ‌‌ and ‌‌‌f′(0+)=1+1=2⇒f′(0−)≠f′(0+) ‌So, it is not differentiable at x=0. (d) We have, ‌f(x)=sin‌|x|−|x|
‌∴f′(x)={
−cos‌x+1
‌
x<0
cos‌x−1
‌
x≥0
. ‌∴f′(0−)=−1+1=−0 ‌‌ and ‌f1(0+)=1−1=0 ‌⇒f′(0−)=f′(0+) ‌ So, it is differentiable at x=0.