Theorem 3 — Derivative of Product of Functions

Theorem 3. If $u$ and $v$ are differentiable functions of $x$ such that $y=uv$, then

Proof. Given $y=uv$. Let $x$ receive a small increment $\delta x$; then $u\to u+\delta u$, $v\to v+\delta v$, $y\to y+\delta y$, so

Subtracting $y=uv$:

Since $\delta x\ne0$, divide throughout by $\delta x$:

Taking the limit as $\delta x\to0$, and using that a differentiable function's own increment $\delta u\to0$ as $\delta x\to0$ (so the last term vanishes):

Since $u,v$ are differentiable, $\lim_{\delta x\to0}\dfrac{\delta u}{\delta x}=\dfrac{du}{dx}$ and $\lim_{\delta x\to0}\dfrac{\delta v}{\delta x}=\dfrac{dv}{dx}$, and $\lim_{\delta x\to0}\delta u=0$; substituting into (1), the last term is $0\cdot\dfrac{dv}{dx}=0$, leaving

Corollary (product of three functions). If $u,v,w$ are differentiable functions of $x$ and $y=uvw$, applying Theorem 3 twice (first to $(uv)$ and $w$, then expanding the derivative of $uv$) gives …