Thank you for reporting, we will resolve it shortly
Q.
Let f(x) = x|x|, g(x) = sin x and h(x) = (gof) (x). Then
Continuity and Differentiability
Solution:
Let f (x) = x|x| = x|x|, g(x) = sin x and h (x) = gof (x) = g[f (x)]
$\therefore \ \ h(x) = \begin{cases}
\sin \ x^2 , & x \geq 0\\
- \sin \ x^2 , & x < 0
\end{cases} $
Now, $h' (x) = \begin{cases}
2x \cos \ x^2 , & x \geq 0\\
-2 x \ \cos \ x^2 , & x < 0
\end{cases} $
Since, L.H.L and R.H.L at x = 0 of h' (x) is equal to 0 therefore h' (x) is continuous at x = 0
Now, suppose h' (x) is differentiable
$\therefore \ \ h''(x) = \begin{cases}
2(\cos x^2 + 2x^2 (-\sin x^2) , & x \geq 0\\
2 (- \cos x^2 + 2x^2 \sin x^2) , & x < 0
\end{cases} $
Since, L.H.L and R.H.L at x = 0 of h" (x) are different therefore h" (x) is not continuous.
$\Rightarrow $ h"(x) is not differentiable
$\Rightarrow $ our assumption is wrong
Hence h'(x) is not differentiable at x = 0.