Derivatives of Products and Quotients

Product Rule

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

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

Examples:

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

Quotient Rule

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

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

Examples:

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

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

Resources

• Product and Quotient Rules, Grover City College
• Product and Quotient Rule, Paul's Online Notes (Lamar University)