$\approx$ ### Theorem The Chain Rule If $f(u)$ is differentiable at the point $u=g(x)$ and $g(x)$ is differentiable at $x$, then the composite function $(f \circ g)(x)=f(g(x))$ is differentiable at $x$, and $ (f \circ g)^{\prime}(x)=f^{\prime}(g(x)) \cdot g^{\prime}(x) $ In Leibniz's notation, if $y=f(u)$ and $u=g(x)$, then $ \frac{d y}{d x}=\frac{d y}{d u} \cdot \frac{d u}{d x} $ ##### Alternate Notation $\begin{aligned} \frac{d y}{d x} &=f^{\prime}(g(x)) \cdot g^{\prime}(x) \\ \frac{d}{d x} f(u) &=f^{\prime}(u) \frac{d u}{d x} \end{aligned}$ #### Proof $\Delta u=g(x+\Delta x)-g(x)\neq0$ $\Delta y=f(u+\Delta u)-f(u)$ $ \begin{aligned} \frac{\Delta y}{\Delta x}&=\frac{\Delta y}{\Delta u} \cdot \frac{\Delta u}{\Delta x}\\ &=\lim _{\Delta x \rightarrow 0} \frac{\Delta y}{\Delta x} \\ \frac{d y}{d x} &=\lim _{\Delta x \rightarrow 0} \frac{\Delta y}{\Delta u} \cdot \frac{\Delta u}{\Delta x} \\ &=\lim _{\Delta x \rightarrow 0} \frac{\Delta y}{\Delta u} \cdot \lim _{\Delta x \rightarrow 0} \frac{\Delta u}{\Delta x} \\ &=\lim _{\Delta u \rightarrow 0} \frac{\Delta y}{\Delta u} \cdot \lim _{\Delta x \rightarrow 0} \frac{\Delta u}{\Delta x} \\ &=\frac{d y}{d u} \cdot \frac{d u}{d x} \end{aligned}$ --- Given $y=f(u)$ and $u=g(x)$ and everything is nice and differentiable. $\Delta x$ is a increment that $\Delta u$ and $\Delta y$ correspond to. Using [[Change in y near x=a]] $ \Delta u=g^{\prime}\left(x_{0}\right) \Delta x+\epsilon_{1} \Delta x=\left(g^{\prime}\left(x_{0}\right)+\epsilon_{1}\right) \Delta x $ where $\epsilon_{1} \rightarrow 0$ as $\Delta x \rightarrow 0 .$ Similarly, $ \Delta y=f^{\prime}\left(u_{0}\right) \Delta u+\epsilon_{2} \Delta u=\left(f^{\prime}\left(u_{0}\right)+\epsilon_{2}\right) \Delta u $ where $\epsilon_{2} \rightarrow 0$ as $\Delta u \rightarrow 0 .$ Notice also that $\Delta u \rightarrow 0$ as $\Delta x \rightarrow 0 .$ Substituting $\Delta u=\left(g^{\prime}\left(x_{0}\right)+\epsilon_{1}\right) \Delta x$ into the equation for $\Delta y$ gives $ \Delta{y}=\left(f^{\prime}\left(u_{0}\right)+\epsilon_{2}\right)\left(g^{\prime}\left(x_{0}\right)+\epsilon_{1}\right) \Delta x $ so, by expanding the binomials, and dividing by $\Delta x$ $\frac{\Delta y}{\Delta x}=f^{\prime}\left(u_{0}\right) g^{\prime}\left(x_{0}\right)+\epsilon_{2} g^{\prime}\left(x_{0}\right)+f^{\prime}\left(u_{0}\right) \epsilon_{1}+\epsilon_{2} \epsilon_{1}$ All the terms with $\epsilon$ will approach 0 $\left.\frac{d y}{d x}\right|_{x=x_{0}}=\lim _{\Delta x \rightarrow 0} \frac{\Delta y}{\Delta x}=f^{\prime}\left(u_{0}\right) g^{\prime}\left(x_{0}\right)=f^{\prime}\left(g\left(x_{0}\right)\right) \cdot g^{\prime}\left(x_{0}\right)$ ## Notes When a function is a composition of more than two functions, the chain rule may need to be applied more than once. To recognize anything other than elementary anti-derivatives, it is necessary to understand the chain rule.