I wanted a little help in proving or disproving the statement: "The function $g(x) = e^{f(x)}$ is always differentiable $\forall~ x \in $ domain of $f(x)$ $~\textbf{if}~$ $f(x)$ is differentiable in it's domain."
$\bullet~$ $\textbf{My approach:}$
I took some examples like $ e^{1/x} , e^{-1/x}, e^{{1}/{x^2}} ,e^{\sin(x)}$ and this statement holds true. I verified them using the first principle. I could not prove it for a general $f(x)$ but if its true then it will save a lot of time which would go to check functions like $g(x) = e^{-1/(x+e) +1/(x-\pi)}$ so, is this statement true? kindly help me out.
Take two functions $f,g:\mathbb{R}\to\mathbb{R}$ and suppose that $f$ is differentiable at $x_0$ and $g$ is differentiable at $f(x_0)$. Formally, this means that the limits $$ \lim_{x\to x_0} \dfrac{f(x) - f(x_0)}{x-x_0}\quad\text{ and }\quad \lim_{y\to f(x_0)} \dfrac{g(y) - g(f(x_0))}{y- f(x_0)} $$ exist. Denote the limits by the usual $f'(x_0)$ and $g'(f(x_0))$. We need to show that the limit $$ \lim_{x\to x_0} \dfrac{g(f(x)) - g(f(x_0))}{x-x_0} $$ exists and is equal to $(g'\circ f)(x_0)\cdot f'(x_0)$.
Intuitively (and naively), one would write for $x\neq x_0$ that $$ \dfrac{g(f(x)) - g(f(x_0))}{x-x_0} = \dfrac{g(f(x)) - g(f(x_0))}{f(x_0) - f(x)}\cdot\dfrac{f(x) - f(x_0)}{x-x_0} $$ and it would suffice to take the limit of both products, because $f$ is necessarily continuous at $x_0$, hence $f(x)\to f(x_0)$ for $x\to x_0$. BUT it is possible that $f(x) = f(x_0)$ for $x\neq x_0$ if $f$ is not injectif. That means that the above doesn't work. And this is the reason why this proof is a bit annoying. So to work around this problem, we define $$ h(x) := \left\{\begin{array}{ll} \dfrac{g(f(x)) - g(f(x_0))}{f(x) - f(x_0)}&\text{ if $f(x) \neq f(x_0)$}\\ g'(f(x_0)) &\text{ if $f(x) = f(x_0)$} \end{array}\right.. $$ Now, this function is well-defined (and we used the most "obvious" definition [if not clear, think about it; if still not clear, write a comment]). Furthermore, it is continuous at $x_0$ (clear? if not, prove it). And that still gives us the result we want, because $$ \dfrac{g(f(x)) - g(f(x_0))}{x-x_0} = h(x)\cdot \dfrac{f(x) - f(x_0)}{x-x_0} \underset{x\to x_0}{\rightarrow} h(x_0)\cdot f'(x_0) $$ and $h(x_0) = g'(f(x_0))$. (Clear, why the above equality holds?)