### Theorem If $u$ and $v$ are differentiable at $x$, then so is their product $u v$, and $ \frac{d}{d x}(u v)=u \frac{d v}{d x}+v \frac{d u}{d x} $ ##### Alternate Notation $(u v)^{\prime}=u v^{\prime}+v u^{\prime}$ $\frac{d}{d x}[f(x) g(x)]=f(x) g^{\prime}(x)+g(x) f^{\prime}(x)$ #### Proof $ \frac{d}{d x}(u v)=\lim _{h \rightarrow 0} \frac{u(x+h) v(x+h)-u(x) v(x)}{h} $ To change this fraction into an equivalent one that contains difference quotients for the derivatives of $u$ and $v$, we subtract and add $u(x+h) v(x)$ in the numerator: $ \begin{aligned} \frac{d}{d x}(u v)&=\lim _{h \rightarrow 0} \frac{u(x+h) v(x+h)-u(x+h) v(x)+u(x+h) v(x)-u(x) v(x)}{h} \\ &=\lim _{h \rightarrow 0}\left[u(x+h) \frac{v(x+h)-v(x)}{h}+v(x) \frac{u(x+h)-u(x)}{h}\right] \\ &=\lim _{h \rightarrow 0} u(x+h) \cdot \lim _{h \rightarrow 0} \frac{v(x+h)-v(x)}{h}+v(x) \cdot \lim _{h \rightarrow 0} \frac{u(x+h)-u(x)}{h} \end{aligned} $ As $h$ approaches zero, $u(x+h)$ approaches $u(x)$ because $u$, being differentiable at $x$, is continuous at $x$. ([[Differentiability implies Continuity]]) The two fractions approach the values of $d v / d x$ at $x$ and $d u / d x$ at $x$. ([[Derivative of a function]] Therefore, $ \frac{d}{d x}(u v)=u \frac{d v}{d x}+v \frac{d u}{d x} $