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 ${\frac {d}{dx}}({\sqrt {x}})(\sin {x})=({\frac {1}{2{\sqrt {x}}}})(\cos {x})+{\sqrt {x}}\sin {x}$.

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

Failed to parse (syntax error): {\displaystyle \frac{d}{dx}\(\frac{f(x)}{g(x)}\) = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2} } .

Resources

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