Chain rule proof - derivative of a composite function

Chain Rule - Derivative of composite function g circle f g ∘ f

Consider $I$ and $J$ two intervals of $\mathbb {R} $ and two functions $ f, g $ defined by

$$ \begin{aligned} f&: I \rightarrow \mathbb{R}\\ g&: J \rightarrow \mathbb{R} \end{aligned} $$

such $f(I) \subset J$. Let $x$ a point of the interval $I$.
If $f$ is differentiable at $x$ and $g$ is differentiable at $f(x)$ then the composite function $g \circ f$ is differentiable at $x$, and the Chaine Rule is given by

$$ \forall x\in I, \quad \left(g \circ f\right)^{\prime}(x)=g^{\prime}(f(x)) \cdot f^{\prime}(x) $$

Proof

We have by definition:

$$ \begin{aligned} \left(g \circ f\right)^{\prime}(x)=&\lim _{h \rightarrow 0} \frac{(g \circ f)(x+h)-(g \circ f)(x)}{h}\\ =&\lim _{h \rightarrow 0} \frac{g(f(x+h))-g(f(x))}{h} \\ =&\lim _{h \rightarrow 0} \frac{g(f(x+h))-g(f(x))}{h} \times \frac{f(x+h)-f(x)}{f(x+h)-f(x)}\\ =&\lim _{h \rightarrow 0} \frac{g(f(x+h))-g(f(x))}{f(x+h)-f(x)} \times \frac{f(x+h)-f(x)}{h}\\ =&\lim _{h \rightarrow 0} \frac{g(f(x+h))-g(f(x))}{f(x+h)-f(x)} \times \lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\\ =&\lim _{h \rightarrow 0} \frac{g(f(x+h))-g(f(x))}{f(x+h)-f(x)} \times f^{\prime}(x) \end{aligned} $$

Using the change of variable $k=f(x+h)-f(x)$

$$ \begin{aligned} k&=f(x+h)-f(x)\\ f(x+h)&=f(x)+k \end{aligned} $$

We have:

$$ \lim _{h \rightarrow 0} k = \lim _{h \rightarrow 0} f(x+h)-f(x) = f(x+0)-f(x)=0 $$

It means, when $h$ approaches 0, then $k$ approaches 0.

Finding the limit when $h$ approaches 0, by using the change of variable, is the same as finding the limit when $k$ approaches 0:

$$ \begin{aligned} \lim _{h \rightarrow 0} \frac{g(f(x+h))-g(f(x))}{f(x+h)-f(x)} =&\lim _{k\rightarrow 0} \frac{g({\color{red}{f(x)}}+k)-g({\color{red}{f(x)}})}{k}\\ &=g^{\prime}({\color{red}{f(x)}}) \end{aligned} $$

We obtain:

$$ \begin{aligned} \left(g \circ f\right)^{\prime}(x)=&\lim _{h \rightarrow 0} \frac{g(f(x+h))-g(f(x))}{f(x+h)-f(x)} \times f^{\prime}(x)\\ &=g^{\prime}(f(x)) \cdot f^{\prime}(x) \end{aligned} $$

We conclude:

$$ \left(g \circ f\right)^{\prime}(x)=g^{\prime}(f(x)) \cdot f^{\prime}(x) $$