Chain rule proof - derivative of a composite function

Derivative of composite function (g ∘ f) (g circle f), chain rule is defined by (g ∘ f)’(x) = g’(f(x)) × f’(x) .

Derivative of composite function (u ∘ v) (u circle v), chain rule is defined by (u ∘ v)’(x) = u’(v(x)) × v’(x) .

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

$$\begin{array}{rl}f& :I\to \mathbb{R}\\ g& :J\to \mathbb{R}\end{array}$$

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

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

Proof

We have by definition:

$$\begin{array}{rl}{(g\circ f)}^{\prime }(x)=& \underset{h\to 0}{lim}\frac{(g\circ f)(x+h)-(g\circ f)(x)}{h}\\ =& \underset{h\to 0}{lim}\frac{g(f(x+h))-g(f(x))}{h}\\ =& \underset{h\to 0}{lim}\frac{g(f(x+h))-g(f(x))}{h}\times \frac{f(x+h)-f(x)}{f(x+h)-f(x)}\\ =& \underset{h\to 0}{lim}\frac{g(f(x+h))-g(f(x))}{f(x+h)-f(x)}\times \frac{f(x+h)-f(x)}{h}\\ =& \underset{h\to 0}{lim}\frac{g(f(x+h))-g(f(x))}{f(x+h)-f(x)}\times \underset{h\to 0}{lim}\frac{f(x+h)-f(x)}{h}\\ =& \underset{h\to 0}{lim}\frac{g(f(x+h))-g(f(x))}{f(x+h)-f(x)}\times f^{\prime }(x)\end{array}$$

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

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

We have:

$$\underset{h\to 0}{lim}k=\underset{h\to 0}{lim}f(x+h)-f(x)=f(x+0)-f(x)=0$$

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:

$$\begin{array}{rl}\underset{h\to 0}{lim}\frac{g(f(x+h))-g(f(x))}{f(x+h)-f(x)}=& \underset{k\to 0}{lim}\frac{g(f(x)+k)-g(f(x))}{k}\\ & =g^{\prime }(f(x))\end{array}$$

We obtain:

$$\begin{array}{rl}{(g\circ f)}^{\prime }(x)=& \underset{h\to 0}{lim}\frac{g(f(x+h))-g(f(x))}{f(x+h)-f(x)}\times f^{\prime }(x)\\ & =g^{\prime }(f(x))\cdot f^{\prime }(x)\end{array}$$

We conclude:

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