The quotient rule is a formula used to find the derivative of a quotient of two functions.
Given two differentiable functions $f(x)$ and $g(x)$ such that $g(x)\neq 0$, the derivative with respect to $x$ of the quotient of $f(x)$ and $g(x)$ is given by:
\[\frac{\mathrm{d} }{\mathrm{d} x}\left[\frac{f(x)}{g(x)}\right]=\frac{f'(x)g(x)-f(x)g'(x)}{[g(x)]^2}.\]
($f'(x)$ denotes the derivative of $f$ with respect to $x$, and $g'(x)$ denotes the derivative of $g$ with respect to $x$.)
Use the quotient rule to show that $\dfrac{\mathrm{d} }{\mathrm{d} x}\left[\tan{x}\right]=\sec^2{x}.$
Recall that $\tan{x}=\dfrac{\sin{x} }{\cos{x} }.$
Let $f(x)=\sin{x}$ and $g(x)=\cos{x}$, then by the quotient rule:
\begin{align} \dfrac{\mathrm{d} }{\mathrm{d} x}\left[\tan{x}\right] &= \dfrac{\mathrm{d} }{\mathrm{d} x}\left[\dfrac{\sin{x} }{\cos{x} }\right] \\ \\ &= \dfrac{\left(\dfrac{\mathrm{d} }{\mathrm{d} x}\left[\sin{x}\right]\right) \cdot \cos{x}-\sin{x} \cdot \left(\dfrac{\mathrm{d} }{\mathrm{d} x}\left[\cos{x}\right]\right)}{\cos^2{x} } \\ \\ &= \dfrac{\cos{x}\cdot\cos{x}-\sin{x}\cdot(-\sin{x})}{\cos^2(x)} \\ \\ &= \dfrac{\cos^2{x}+\sin^2{x} }{\cos^2{x} } \end{align}
This can be simplified using the trigonometric identity $\cos^2{x}+\sin^2{x}=1$:
\[\dfrac{\mathrm{d} }{\mathrm{d} x}\left[\tan{x}\right] = \dfrac{1}{\cos^2{x} }.\]
And recall that $\sec{x}=\dfrac{1}{\cos{x} }$, so:
\[\dfrac{\mathrm{d} }{\mathrm{d} x}\left[\tan{x}\right] = \sec^2{x}.\]
Prof. Robin Johnson differentiates $\dfrac{1+x^3}{1+x^2}$.
Test yourself: Numbas test on the quotient rule