I am studying advanced calculus and I came across this problem. (The book is Advanced Calculus second edition by Patrick M. Fitzpatrick.)
Let $I$ and $J$ be open intervals, and the functions $f:I\rightarrow\mathbb{R}$ and $h:J\rightarrow\mathbb{R}$ have the property that $h(J)\subseteq I$, so the composition $f\circ g:J\rightarrow\mathbb{R}$ is defined. Show that if $x_0$ is in $J$, $h:J\rightarrow\mathbb{R}$ is continuous at $x_0$, $h(x)\neq h(x_0)$ if $x\neq x_0$, and $f:I\rightarrow\mathbb{R}$ is differentiable at $h(x_0)$, then $$lim_{x\rightarrow x_0}\frac{f(h(x))-f(h(x_0))}{h(x)-h(x_0)}=f'(h(x_0))$$.
I understand that the question is saying show and not prove which usually means that it is obvious but it already said that $f$ was differentable and so by definition $f'$ exists and furthermore it states that $h(J)\subseteq I$ so that $f$ is clearly defined for any $h(x)$. What is left to show. It seems like they are just rewriting the definition.
We can sketch a solution here using the $\epsilon$-$\delta$ definition of limits. Our goal is to show that for every $\epsilon$, there exists $\delta > 0$ such that
$$ x \in J \setminus \{x_0\} \text{ and } |x - x_0| < \delta \implies \left|\frac{f(h(x)) - f(h(x_0))}{h(x) - h(x_0)} - f'(h(x_0))\,\right| < \epsilon \;. $$
Thus, let $\epsilon > 0$ be given. $f$ is differentiable, so there exists $\delta_1 > 0$ such that
$$ y \in I \,\setminus\, \{h(x_0)\} \text{ and } |y - h(x_0)| < \delta_1 \implies \left|\frac{f(y) - f(h(x_0))}{y - h(x_0)} - f'(h(x_0))\,\right| < \epsilon \;. $$
Since $h$ is continuous at $x_0$, there exists $\delta_2 > 0$ such that
$$ x \in J \text{ and } |x - x_0| < \delta_2 \implies |h(x) - h(x_0)| < \delta_1 \;. $$
Note that $h(x) \neq h(x_0)$ for $x \neq x_0$, so $h$ is injective. Then for any $x \in J \setminus \{x_0\}$, we also have $h(x) \in h(J) \setminus \{h(x_0)\} \subset I \setminus \{h(x_0)\}$.
Putting $\delta :=\delta_2$, we satisfy the above.