MA005: Calculus I

Chain Rule (Leibniz notation form)

If $y$ is a differentiable function of $u$, and $u$ is a differentiable function of $x$, then $y$ is a differentiable function of $x$ and $\frac{d y}{d x}=\frac{d y}{d u} \cdot \frac{d u}{d x}$.

Idea for a proof: $\frac{\mathrm{dy}}{\mathrm{dx}}=\lim _{\Delta x \rightarrow 0} \frac{\Delta y}{\Delta u} \cdot \frac{\Delta u}{\Delta x} \quad$ (if $\left.\Delta u \neq 0\right)$

$\begin{align*}\begin{aligned}&=\left(\lim _{\Delta x \rightarrow 0} \frac{\Delta y}{\Delta u}\right) \cdot\left(\lim _{\Delta x \rightarrow 0} \frac{\Delta u}{\Delta x}\right) \quad u \text { is continuous, so } \Delta x \rightarrow 0 \text { implies } \Delta u \rightarrow 0 \\&=\frac{d y}{d u} \cdot \frac{d u}{d x}\end{aligned}\end{align*}$

Although this nice short argument gets to the heart of why the Chain Rule works, it is not quite valid. If $\mathrm{du} / \mathrm{d} \mathrm{x} \neq 0$, then it is possible to show that $\Delta \mathrm{u} \neq 0$ for all very small values of $\Delta \mathrm{x}$, and the "idea for a proof" is a real proof. There are, however, functions for which $\Delta \mathrm{u}=0$ for lots of small values of $\Delta \mathrm{x}$, and these create problems for the previous argument. A justification which is true for ALL cases is much more complicated.

The symbol $\frac{\mathrm{dy}}{\mathrm{du}}$ is a single symbol ( as is $\frac{\mathrm{du}}{\mathrm{dx}}$ ), and we cannot eliminate du from the product $\frac{\mathrm{dy}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$ in the Chain Rule. It is, however, perfectly fine to use the idea of eliminating du to help you remember the statement of the Chain Rule.

Example 3: $\quad y=\cos \left(x^{2}+3\right)$ is $y=\cos (u)$ with $u=x^{2}+3$. Find $d y / d x$

Solution: $\mathrm{y}=\cos (\mathrm{u})$ so $\mathrm{dy} / \mathrm{du}=-\sin (\mathrm{u}) . \mathrm{u}=\mathrm{x}^{2}+3$ so $\mathrm{du} / \mathrm{dx}=2 \mathrm{x}$. Finally, using the Chain Rule, $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dy}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}=-\sin (\mathrm{u}) \cdot 2 \mathrm{x}=-2 \mathrm{x} \cdot \sin \left(\mathrm{x}^{2}+3\right)$

Practice 2: Find $\mathrm{dy} / \mathrm{dx}$ for $\mathrm{y}=\sin \left(4 \mathrm{x}+\mathrm{e}^{\mathrm{X}}\right)$

There is also a composition of functions form of the Chain Rule. The notation is different, but it means precisely the same as the Leibniz form.