Difference between revisions of "Derivatives of Products and Quotients"
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Examples: | Examples: | ||
* <math> f(x) = 6x + 1 </math>, <math> g(x) = x^2 </math>. <math> f'(x) = 6 </math> and <math> g'(x) = 2x </math>, so <math> \frac{d}{dx}(6x + 1)(x^2) = 6x^2 + (2x)(6x + 1)</math>. | * <math> f(x) = 6x + 1 </math>, <math> g(x) = x^2 </math>. <math> f'(x) = 6 </math> and <math> g'(x) = 2x </math>, so <math> \frac{d}{dx}(6x + 1)(x^2) = 6x^2 + (2x)(6x + 1)</math>. | ||
− | * <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>. | + | * <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}) = \left(\frac{1}{2\sqrt{x}}\right)(\cos{x}) + \sqrt{x}\sin{x}</math>. |
===Quotient Rule=== | ===Quotient Rule=== |
Revision as of 10:34, 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
.
Resources
- Product and Quotient Rules, Grover City College
- Product and Quotient Rule, Paul's Online Notes (Lamar University)