 Quotient Rule
Get Quotient Rule essential facts below. View Videos or join the Quotient Rule discussion. Add Quotient Rule to your PopFlock.com topic list for future reference or share this resource on social media.
Quotient Rule

In calculus, the quotient rule is a method of finding the derivative of a function that is the ratio of two differentiable functions. Let $f(x)=g(x)/h(x),$ where both $g$ and $h$ are differentiable and $h(x)\neq 0.$ The quotient rule states that the derivative of $f(x)$ is

$f'(x)={\frac {g'(x)h(x)-g(x)h'(x)}{[h(x)]^{2}}}.$ ## Examples

1. A basic example:
{\begin{aligned}{\frac {d}{dx}}{\frac {e^{x}}{x^{2}}}&={\frac {\left({\frac {d}{dx}}e^{x}\right)(x^{2})-(e^{x})\left({\frac {d}{dx}}x^{2}\right)}{(x^{2})^{2}}}\\&={\frac {(e^{x})(x^{2})-(e^{x})(2x)}{x^{4}}}\\&={\frac {e^{x}(x-2)}{x^{3}}}.\end{aligned}} 2. The quotient rule can be used to find the derivative of $f(x)=\tan x={\tfrac {\sin x}{\cos x}}$ as follows.
{\begin{aligned}{\frac {d}{dx}}\tan x&={\frac {d}{dx}}{\frac {\sin x}{\cos x}}\\&={\frac {\left({\frac {d}{dx}}\sin x\right)(\cos x)-(\sin x)\left({\frac {d}{dx}}\cos x\right)}{\cos ^{2}x}}\\&={\frac {\cos ^{2}x+\sin ^{2}x}{\cos ^{2}x}}\\&={\frac {1}{\cos ^{2}x}}=\sec ^{2}x.\end{aligned}} ## Proofs

### Proof from derivative definition and limit properties

Let $f(x)=g(x)/h(x).$ Applying the definition of the derivative and properties of limits gives the following proof.

{\begin{aligned}f'(x)&=\lim _{k\to 0}{\frac {f(x+k)-f(x)}{k}}\\&=\lim _{k\to 0}{\frac {{\frac {g(x+k)}{h(x+k)}}-{\frac {g(x)}{h(x)}}}{k}}\\&=\lim _{k\to 0}{\frac {g(x+k)h(x)-g(x)h(x+k)}{k\cdot h(x)h(x+k)}}\\&=\lim _{k\to 0}{\frac {g(x+k)h(x)-g(x)h(x+k)}{k}}\cdot \lim _{k\to 0}{\frac {1}{h(x)h(x+k)}}\\&=\left(\lim _{k\to 0}{\frac {g(x+k)h(x)-g(x)h(x)+g(x)h(x)-g(x)h(x+k)}{k}}\right)\cdot {\frac {1}{h(x)^{2}}}\\&=\left(\lim _{k\to 0}{\frac {g(x+k)h(x)-g(x)h(x)}{k}}-\lim _{k\to 0}{\frac {g(x)h(x+k)-g(x)h(x)}{k}}\right)\cdot {\frac {1}{h(x)^{2}}}\\&=\left(h(x)\lim _{k\to 0}{\frac {g(x+k)-g(x)}{k}}-g(x)\lim _{k\to 0}{\frac {h(x+k)-h(x)}{k}}\right)\cdot {\frac {1}{h(x)^{2}}}\\&={\frac {g'(x)h(x)-g(x)h'(x)}{h(x)^{2}}}.\end{aligned}} ### Proof using implicit differentiation

Let $f(x)={\frac {g(x)}{h(x)}},$ so $g(x)=f(x)h(x).$ The product rule then gives $g'(x)=f'(x)h(x)+f(x)h'(x).$ Solving for $f'(x)$ and substituting back for $f(x)$ gives:

{\begin{aligned}f'(x)&={\frac {g'(x)-f(x)h'(x)}{h(x)}}\\&={\frac {g'(x)-{\frac {g(x)}{h(x)}}\cdot h'(x)}{h(x)}}\\&={\frac {g'(x)h(x)-g(x)h'(x)}{h(x)^{2}}}.\end{aligned}} ### Proof using the chain rule

Let $f(x)={\frac {g(x)}{h(x)}}=g(x)h(x)^{-1}.$ Then the product rule gives

$f'(x)=g'(x)h(x)^{-1}+g(x)\cdot {\frac {d}{dx}}(h(x)^{-1}).$ To evaluate the derivative in the second term, apply the power rule along with the chain rule:

$f'(x)=g'(x)h(x)^{-1}+g(x)\cdot (-1)h(x)^{-2}h'(x).$ Finally, rewrite as fractions and combine terms to get

{\begin{aligned}f'(x)&={\frac {g'(x)}{h(x)}}-{\frac {g(x)h'(x)}{h(x)^{2}}}\\&={\frac {g'(x)h(x)-g(x)h'(x)}{h(x)^{2}}}.\end{aligned}} ## Higher order formulas

Implicit differentiation can be used to compute the nth derivative of a quotient (partially in terms of its first n − 1 derivatives). For example, differentiating $fh=g$ twice (resulting in $f''h+2f'h'+fh''=g''$ ) and then solving for $f''$ yields

$f''=\left({\frac {g}{h}}\right)''={\frac {g''-2f'h'-fh''}{h}}.$ 