Difference between revisions of "Derivatives of Products and Quotients"
Jump to navigation
Jump to search
(2 intermediate revisions by the same user not shown) | |||
Line 6: | Line 6: | ||
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=== | ||
Line 12: | Line 12: | ||
<math> \frac{d}{dx}\left(\frac{f(x)}{g(x)}\right) = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2} </math>. | <math> \frac{d}{dx}\left(\frac{f(x)}{g(x)}\right) = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2} </math>. | ||
+ | |||
+ | 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>. | ||
+ | |||
==Resources== | ==Resources== | ||
* [http://www2.gcc.edu/dept/math/faculty/BancroftED/buscalc/chapter2/section2-4.php Product and Quotient Rules], Grover City College | * [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) | * [https://tutorial.math.lamar.edu/classes/calci/productquotientrule.aspx Product and Quotient Rule], Paul's Online Notes (Lamar University) |
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)