Difference between revisions of "Derivatives of Products and Quotients"

Revision as of 10:31, 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 ${\frac {d}{dx}}({\sqrt {x}})(\sin {x})=({\frac {1}{2{\sqrt {x}}}})(\cos {x})+{\sqrt {x}}\sin {x}$ .

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

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} } .

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 * <math> f(x) = \sqrt{x} </math>, <math> g(x) = \sin{x} </math>. <math> f'(x) = \frac{1}{2\sqrt{x}} </math> and <math> g'(x) = \cos{x} </math>, so <math> \frac{d}{dx}(\sqrt{x})(\sin{x}) = (\frac{1}{2\sqrt{x}})(\cos{x}) + \sqrt{x}\sin{x}</math>.
+===Quotient Rule===
+Let <math> f(x) </math> and <math> g(x) </math> be differentiable functions such that <math> g(x) \neq 0 </math>. Then, the derivative of the quotient, <math> \frac{f(x)}{g(x)} </math>, is
-===Quotient Rule===
+<math> \frac{d}{dx}\(\frac{f(x)}{g(x)}\) = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2} </math>.
 ==Resources==
 * [http://www2.gcc.edu/dept/math/faculty/BancroftED/buscalc/chapter2/section2-4.php Product and Quotient Rules], Grover City College
 * [https://tutorial.math.lamar.edu/classes/calci/productquotientrule.aspx Product and Quotient Rule], Paul's Online Notes (Lamar University)