If the composition function $g(f(x))$ is differentiable, is $f(x)$ differentiable?

Question

I am trying to prove that the function $f_a(x) = e^{-\frac{1}{(x-a)}}$ is differentiable for all $x>a$. However, I do not know how to show $|\frac{e^{-\frac{1}{(x-a)}}-e^{-\frac{1}{(p-a)}}}{x-p} - (-\frac{e^{-\frac{1}{(p-a)}}}{(p-a)^2})| < \epsilon$ if $0<|x-p|<\delta$.

My next idea was to apply the natural logarithmic function to $e^{-\frac{1}{(x-a)}}$ to get $\hat{f_a}(x)=-\frac{1}{(x-a)}$. From here, I can prove that $\hat{f_a}(x)$ is differentiable for any $x>a$, but I do not know if this implies that $f_a(x)$ is differentiable for any $x>a$.

I know that $f_a(x):(-\infty,\infty) \rightarrow (0,\infty)$ and $\hat{f_a}(x):(-\infty,\infty) \rightarrow (-\infty,\infty)$.

Can you please help me understand how I can prove $f_a(x)$ is differentiable for all $x>a$? Thank you for your time and help.

Answer 1

There are plenty of proofs for the chain rule, that is if $f(x)$ and $g(x)$ are differentiable, then so is $g(f(x))$. You can take it upon yourself to verify this, as it is typically standard for students to prove these statements they use every day in calculus. As such you can find proofs of such, I believe I have answered one before. Then $\frac{d}{dx} e^x= e^x$ and $\frac{d}{dx} \frac{-1}{a-x}=\frac{-1}{(a-x)^2}$ consider now this is defined for all $x \neq a $ because the denominator is zero.

To discuss how to prove this, you can seek to prove it directly or you can prove greater generality about all functions and composition of functions, then plug this application into that.

Answer 2

You were almost there! Note that $f_{a}: x \overset{g}{\mapsto} \frac{-1}{x-a} =: y \overset{h}{\mapsto} e^{y}$. You also know that both $g$ and $h$ are differentiable. So by the chain rule $f_{a} = h \circ g$ is also differentiable (its proof is contained as a special case in the proof of chain rule).

The comments above already gave some great examples for your reference. Let me supplement something that may help you conceptually. A theorem of the form "if $P$ then $Q$" is not applicable when $P$ does not hold.