Difference between revisions of "Derivatives of Products and Quotients"
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+ | * <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>. | ||
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* <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>. | * <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>. | ||
Latest revision as of 12:37, 21 September 2021
Product Rule
Let and be differentiable functions. Then, the derivative of their product, , is
.
Examples:
- , . and , so .
- , . and , so .
Quotient Rule
Let and be differentiable functions such that . Then, the derivative of the quotient, , is
.
Examples:
- , . and , so .
- , . and , so .
Resources
- Product and Quotient Rules, Grover City College
- Product and Quotient Rule, Paul's Online Notes (Lamar University)