Let g: ℝ → ℝ be a differentiable function with g(0) = 0, g ^ ′ (0)...

f\left(x\right) = \left\{\begin{cases}g\left(x\right) & x > 0 \\ 0 & x = 0 \\ - g\left(x\right) & x < 0\end{cases}\right) <br >LHD: \lim_{x \to {0}^{-}} \left(- {g}^{′}\left(x\right)\right) = - \lim_{x \to {0}^{-}} {g}^{′}\left(x\right) = - 0 = 0$<br >$RHD: \lim_{x \to {0}^{+}} {g}^{′}\left(x\right) = {g}^{′}\left({0}^{+}\right) = 0$<br >Hence$f\left(x\right)$is differentiable at$x = 0$<br >$h\left(x\right) = {e}^{\left|x\right|} = \left\{$\begin{cases}{e}^{x} & x \ge 0 \\ {e}^{- x} & x < 0\end{cases}$\right)
$LHD: \lim_{x \to {0}^{-}} \left(- {e}^{- x}\right) = - 1$
$RHD: \lim_{x \to {0}^{+}} \left({e}^{x}\right) = 1$
Hence $h\left(x\right)$ is not differentiable at $x = 0$
f\left(h\left(x\right)\right) = \left\{\dfrac{{e}^{\left|x\right|}}{\left|{e}^{\left|x\right|}\right|}\rightg\left({e}^{\left|x\right|}\right)
f\left(h\left(x\right)\right) = \left\{\begin{cases}g\left({e}^{x}\right) & x \ge 0 \\ g\left({e}^{- x}\right) & x < 0\end{cases}\right) <br >LHD: \lim_{x \to {0}^{+}} {g}^{′}\left({e}^{- x}\right) \times - {e}^{- x} = {g}^{′}\left(1\right) \times - 1 = - {g}^{′}\left(1\right)$<br >$RHD: \lim_{x \to {0}^{+}} {g}^{′}\left({e}^{x}\right) \times{e}^{x} = {g}^{′}\left(1\right)$<br >Hence foh is non-differentiable at$x = 0$<br >$h\left(f\left(x\right)\right) = \left\{$\begin{cases}1 & x = 0 \\ {e}^{\left|\dfrac{x}{\left|x\right|}g\left(x\right)\right|} & x \ne 0\end{cases}$\right)
= \left\{\begin{cases}1 & x = 0 \\ {e}^{\left|g\left(x\right)\right|} & x \ne 0\end{cases}\right) <br >LHD: \lim_{x \to {0}^{-}} {e}^{\left|g\left(x\right)\right|} \times \pm {g}^{′}\left(x\right) = {e}^{\left|g\left({0}^{-}\right)\right|} \times \pm \left({g}^{′}\left({0}^{-}\right)\right) = 0$<br >$RHD: \lim_{x \to {0}^{+}} {e}^{\left|g\left(x\right)\right|} \times \pm {g}^{′}\left(x\right) = {e}^{\left|g\left({0}^{+}\right)\right|} \times \pm {g}^{′}\left({0}^{+}\right) = 0