Difference between revisions of "Derivatives of Products and Quotients"

Latest revision as of 12:37, 21 September 2021

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

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 ${\frac {d}{dx}}\left({\frac {5x}{x^{2}+1}}\right)={\frac {5(x^{2}+1)-5x(2x)}{(x^{2}+1)^{2}}}={\frac {5x^{2}+5-10x^{2}}{(x^{2}+1)^{2}}}={\frac {5(1-x^{2})}{(x^{2}+1)^{2}}}$ .

@@ Line 14: / Line 14: @@
 Examples:
+* <math> f(x) = x + 1 </math>, <math> g(x) = x^3 </math>. <math> f'(x) = 1 </math> and <math> g'(x) = 3x^2 </math>, so <math> \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}</math>.
 * <math> f(x) = 5x </math>, <math> g(x) = x^2 + 1 </math>. <math> f'(x) = 5 </math> and <math> g'(x) = 2x </math>, so <math> \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}</math>.