# Derivatives of Products and Quotients

### Product Rule

Let ${\displaystyle f(x)}$ and ${\displaystyle g(x)}$ be differentiable functions. Then, the derivative of their product, ${\displaystyle f(x)g(x)}$, is

${\displaystyle {\frac {d}{dx}}(f(x)g(x))=f'(x)g(x)+f(x)g'(x)}$.

Examples:

• ${\displaystyle f(x)=6x+1}$, ${\displaystyle g(x)=x^{2}}$. ${\displaystyle f'(x)=6}$ and ${\displaystyle g'(x)=2x}$, so ${\displaystyle {\frac {d}{dx}}(6x+1)(x^{2})=6x^{2}+(2x)(6x+1)}$.
• ${\displaystyle f(x)={\sqrt {x}}}$, ${\displaystyle g(x)=\sin {x}}$. ${\displaystyle f'(x)={\frac {1}{2{\sqrt {x}}}}}$ and ${\displaystyle g'(x)=\cos {x}}$, so ${\displaystyle {\frac {d}{dx}}({\sqrt {x}})(\sin {x})=\left({\frac {1}{2{\sqrt {x}}}}\right)(\cos {x})+{\sqrt {x}}\sin {x}}$.

### Quotient Rule

Let ${\displaystyle f(x)}$ and ${\displaystyle g(x)}$ be differentiable functions such that ${\displaystyle g(x)\neq 0}$. Then, the derivative of the quotient, ${\displaystyle {\frac {f(x)}{g(x)}}}$, is

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

Examples:

• ${\displaystyle f(x)=x+1}$, ${\displaystyle g(x)=x^{3}}$. ${\displaystyle f'(x)=1}$ and ${\displaystyle g'(x)=3x^{2}}$, so ${\displaystyle {\frac {d}{dx}}\left({\frac {x+1}{x^{3}}}\right)={\frac {x^{3}-(x+1)(3x^{2})}{(x^{3})^{2}}}={\frac {x^{3}-3x^{3}-3x^{2}}{x^{6}}}={\frac {-x^{2}(2x+3)}{x^{6}}}=-{\frac {2x+3}{x^{4}}}}$.
• ${\displaystyle f(x)=5x}$, ${\displaystyle g(x)=x^{2}+1}$. ${\displaystyle f'(x)=5}$ and ${\displaystyle g'(x)=2x}$, so ${\displaystyle {\frac {d}{dx}}\left({\frac {5x}{x^{2}+1}}\right)={\frac {5(x^{2}+1)-5x(2x)}{(x^{2}+1)^{2}}}={\frac {5x^{2}+5-10x^{2}}{(x^{2}+1)^{2}}}={\frac {5(1-x^{2})}{(x^{2}+1)^{2}}}}$.