### Theorem If $u$ and $v$ are differentiable at $x$ and if $v(x) \neq 0$, then the quotient $u / v$ is differentiable at $x$, and $\frac{d}{d x}\left(\frac{u}{v}\right)=\frac{v \frac{d u}{d x}-u \frac{d v}{d x}}{v^{2}}$ ##### Alternate Notation $\left(\frac{u}{v}\right)^{\prime} = \frac{u^\prime v - v^\prime u}{v^2} $ $\frac{d}{d x}\left[\frac{f(x)}{g(x)}\right]=\frac{g(x) f^{\prime}(x)-f(x) g^{\prime}(x)}{g^{2}(x)}$ #### Proof $ \begin{aligned} \frac{d}{d x}\left(\frac{u}{v}\right) &=\lim _{h \rightarrow 0} \frac{\frac{u(x+h)}{v(x+h)}-\frac{u(x)}{v(x)}}{h} \\ &=\lim _{h \rightarrow 0} \frac{v(x) u(x+h)-u(x) v(x+h)}{h v(x+h) v(x)} \end{aligned} $ To change the last fraction into an equivalent one that contains the difference quotients for the derivatives of $u$ and $v$, we subtract and add $v(x) u(x)$ in the numerator. We then get $\begin{aligned} \frac{d}{d x}\left(\frac{u}{v}\right) &=\lim _{h \rightarrow 0} \frac{v(x) u(x+h)-v(x) u(x)+v(x) u(x)-u(x) v(x+h)}{h v(x+h) v(x)} \\ &=\lim _{h \rightarrow 0} \frac{v(x) \frac{u(x+h)-u(x)}{h}-u(x) \frac{v(x+h)-v(x)}{h}}{v(x+h) v(x)}\\ &=\frac{v \frac{d u}{d x}-u \frac{d v}{d x}}{v^{2}} \end{aligned}$