Theorem: Derivative of a Composite Function (the Chain Rule)

Theorem (Chain Rule). If $y=f(u)$ is a differentiable function of $u$, and $u=g(x)$ is a differentiable function of $x$, then $y$ is a differentiable function of $x$, and $\dfrac{dy}{dx}=\dfrac{dy}{du}\cdot\dfrac{du}{dx}.$

Proof (using increments). Let $x$ change by a small amount $\delta x$ (assume $\delta x\ne0$). This produces a corresponding small change $\delta u$ in $u=g(x)$, which in turn produces a corresponding small change $\delta y$ in $y=f(u)$. Assuming $u$ is not locally constant so $\delta u\ne0$, we can write the exact algebraic identity $\dfrac{\delta y}{\delta x}=\dfrac{\delta y}{\delta u}\cdot\dfrac{\delta u}{\delta x}.$ Since $u=g(x)$ is continuous (being differentiable), as $\delta x\to0$ we also get $\delta u\to0$. Taking the limit $\delta x\to0$ on both sides, the right side becomes $\left(\displaystyle\lim_{\delta u\to0}\dfrac{\delta y}{\delta u}\right)\left(\displaystyle\lim_{\delta x\to0}\dfrac{\delta u}{\delta x}\right)=\dfrac{dy}{du}\cdot\dfrac{du}{dx}$, both of which exist and are finite because $y$ is differentiable in $u$ and $u$ is differentiable in $x$. Since the right-hand side has a finite limit, the left-hand side $\displaystyle\lim_{\delta x\to0}\dfrac{\delta y}{\delta x}=\dfrac{dy}{dx}$ must exist too, and equals that same product. This proves $y$ is differentiable in $x$ with $\dfrac{dy}{dx}=\dfrac{dy}{du}\cdot\dfrac{du}{dx}$.

Equivalent Leibniz form. Writing the composite directly as $y=f[g(x)]$, the same result reads $\dfrac{d}{dx}f[g(x)]=f'[g(x)]\cdot g'(x)$: differentiate the outer function $f$ at the point $g(x)$, then multiply by the derivative of the inner function $g$. …