Example 1: Basic example

Given ${\displaystyle h(x)={\frac {e^{x}}{x^{2}}}}$, let ${\displaystyle f(x)=e^{x},g(x)=x^{2}}$, then using the quotient rule:

$${\displaystyle {\begin{aligned}{\frac {d}{dx}}\left({\frac {e^{x}}{x^{2}}}\right)&={\frac {\left({\frac {d}{dx}}e^{x}\right)(x^{2})-(e^{x})\left({\frac {d}{dx}}x^{2}\right)}{(x^{2})^{2}}}\\&={\frac {(e^{x})(x^{2})-(e^{x})(2x)}{x^{4}}}\\&={\frac {x^{2}e^{x}-2xe^{x}}{x^{4}}}\\&={\frac {xe^{x}-2e^{x}}{x^{3}}}\\&={\frac {e^{x}(x-2)}{x^{3}}}.\end{aligned}}}$$

Example 2: Derivative of tangent function

The quotient rule can be used to find the derivative of ${\displaystyle \tan x={\frac {\sin x}{\cos x}}}$ as follows:

$${\displaystyle {\begin{aligned}{\frac {d}{dx}}\tan x&={\frac {d}{dx}}\left({\frac {\sin x}{\cos x}}\right)\\&={\frac {\left({\frac {d}{dx}}\sin x\right)(\cos x)-(\sin x)\left({\frac {d}{dx}}\cos x\right)}{\cos ^{2}x}}\\&={\frac {(\cos x)(\cos x)-(\sin x)(-\sin x)}{\cos ^{2}x}}\\&={\frac {\cos ^{2}x+\sin ^{2}x}{\cos ^{2}x}}\\&={\frac {1}{\cos ^{2}x}}=\sec ^{2}x.\end{aligned}}}$$

Proof from derivative definition and limit properties

Let ${\displaystyle h(x)={\frac {f(x)}{g(x)}}.}$ Applying the definition of the derivative and properties of limits gives the following proof, with the term ${\displaystyle f(x)g(x)}$ added and subtracted to allow splitting and factoring in subsequent steps without affecting the value:

$${\displaystyle {\begin{aligned}h'(x)&=\lim _{k\to 0}{\frac {h(x+k)-h(x)}{k}}\\&=\lim _{k\to 0}{\frac {{\frac {f(x+k)}{g(x+k)}}-{\frac {f(x)}{g(x)}}}{k}}\\&=\lim _{k\to 0}{\frac {f(x+k)g(x)-f(x)g(x+k)}{k\cdot g(x)g(x+k)}}\\&=\lim _{k\to 0}{\frac {f(x+k)g(x)-f(x)g(x+k)}{k}}\cdot \lim _{k\to 0}{\frac {1}{g(x)g(x+k)}}\\&=\lim _{k\to 0}\left[{\frac {f(x+k)g(x)-f(x)g(x)+f(x)g(x)-f(x)g(x+k)}{k}}\right]\cdot {\frac {1}{g(x)^{2}}}\\&=\left[\lim _{k\to 0}{\frac {f(x+k)g(x)-f(x)g(x)}{k}}-\lim _{k\to 0}{\frac {f(x)g(x+k)-f(x)g(x)}{k}}\right]\cdot {\frac {1}{g(x)^{2}}}\\&=\left[\lim _{k\to 0}{\frac {f(x+k)-f(x)}{k}}\cdot g(x)-f(x)\cdot \lim _{k\to 0}{\frac {g(x+k)-g(x)}{k}}\right]\cdot {\frac {1}{g(x)^{2}}}\\&={\frac {f'(x)g(x)-f(x)g'(x)}{g(x)^{2}}}.\end{aligned}}}$$

The limit evaluation ${\displaystyle \lim _{k\to 0}{\frac {1}{g(x+k)g(x)}}={\frac {1}{g(x)^{2}}}}$ is justified by the differentiability of ${\displaystyle g(x)}$, implying continuity, which can be expressed as ${\displaystyle \lim _{k\to 0}g(x+k)=g(x)}$.

Proof using implicit differentiation

Let ${\displaystyle h(x)={\frac {f(x)}{g(x)}},}$ so that ${\displaystyle f(x)=g(x)h(x).}$

The product rule then gives ${\displaystyle f'(x)=g'(x)h(x)+g(x)h'(x).}$

Solving for ${\displaystyle h'(x)}$ and substituting back for ${\displaystyle h(x)}$ gives:

$${\displaystyle {\begin{aligned}h'(x)&={\frac {f'(x)-g'(x)h(x)}{g(x)}}\\&={\frac {f'(x)-g'(x)\cdot {\frac {f(x)}{g(x)}}}{g(x)}}\\&={\frac {f'(x)g(x)-f(x)g'(x)}{g(x)^{2}}}.\end{aligned}}}$$

Proof using the reciprocal rule or chain rule

Let ${\displaystyle h(x)={\frac {f(x)}{g(x)}}=f(x)\cdot {\frac {1}{g(x)}}.}$

Then the product rule gives ${\displaystyle h'(x)=f'(x)\cdot {\frac {1}{g(x)}}+f(x)\cdot {\frac {d}{dx}}\left[{\frac {1}{g(x)}}\right].}$

To evaluate the derivative in the second term, apply the reciprocal rule, or the power rule along with the chain rule:

$${\displaystyle {\frac {d}{dx}}\left[{\frac {1}{g(x)}}\right]=-{\frac {1}{g(x)^{2}}}\cdot g'(x)={\frac {-g'(x)}{g(x)^{2}}}.}$$

Substituting the result into the expression gives

$${\displaystyle {\begin{aligned}h'(x)&=f'(x)\cdot {\frac {1}{g(x)}}+f(x)\cdot \left[{\frac {-g'(x)}{g(x)^{2}}}\right]\\&={\frac {f'(x)}{g(x)}}-{\frac {f(x)g'(x)}{g(x)^{2}}}\\&={\frac {g(x)}{g(x)}}\cdot {\frac {f'(x)}{g(x)}}-{\frac {f(x)g'(x)}{g(x)^{2}}}\\&={\frac {f'(x)g(x)-f(x)g'(x)}{g(x)^{2}}}.\end{aligned}}}$$

Proof by logarithmic differentiation

Let ${\displaystyle h(x)={\frac {f(x)}{g(x)}}.}$ Taking the absolute value and natural logarithm of both sides of the equation gives

$${\displaystyle \ln |h(x)|=\ln \left|{\frac {f(x)}{g(x)}}\right|}$$

Applying properties of the absolute value and logarithms,

$${\displaystyle \ln |h(x)|=\ln |f(x)|-\ln |g(x)|}$$

Taking the logarithmic derivative of both sides,

$${\displaystyle {\frac {h'(x)}{h(x)}}={\frac {f'(x)}{f(x)}}-{\frac {g'(x)}{g(x)}}}$$

Solving for ${\displaystyle h'(x)}$ and substituting back ${\displaystyle {\tfrac {f(x)}{g(x)}}}$ for ${\displaystyle h(x)}$ gives:

$${\displaystyle {\begin{aligned}h'(x)&=h(x)\left[{\frac {f'(x)}{f(x)}}-{\frac {g'(x)}{g(x)}}\right]\\&={\frac {f(x)}{g(x)}}\left[{\frac {f'(x)}{f(x)}}-{\frac {g'(x)}{g(x)}}\right]\\&={\frac {f'(x)}{g(x)}}-{\frac {f(x)g'(x)}{g(x)^{2}}}\\&={\frac {f'(x)g(x)-f(x)g'(x)}{g(x)^{2}}}.\end{aligned}}}$$

Taking the absolute value of the functions is necessary for the logarithmic differentiation of functions that may have negative values, as logarithms are only real-valued for positive arguments. This works because ${\displaystyle {\tfrac {d}{dx}}(\ln |u|)={\tfrac {u'}{u}}}$, which justifies taking the absolute value of the functions for logarithmic differentiation.