Difference between revisions of "Product rule"
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==Definition== | ==Definition== | ||
Given two functions {{M|f:\mathbb{R}\rightarrow\mathbb{R} }} and {{M|g:\mathbb{R}\rightarrow\mathbb{R} }} which are differentiable (at {{M|p}}) the composite function {{M|h:\mathbb{R}\rightarrow\mathbb{R} }} where {{M|1=h=fg}} has derivative: | Given two functions {{M|f:\mathbb{R}\rightarrow\mathbb{R} }} and {{M|g:\mathbb{R}\rightarrow\mathbb{R} }} which are differentiable (at {{M|p}}) the composite function {{M|h:\mathbb{R}\rightarrow\mathbb{R} }} where {{M|1=h=fg}} has derivative: | ||
* <math>\frac{dh}{dx}\Bigg|_p=\frac{d}{dx}[fg]\Bigg|_p=f(p)\frac{dg}{dx}\Bigg|_p+g(p)\frac{df}{dx}\Bigg|_p</math> | * <math>\frac{dh}{dx}\Bigg|_p=\frac{d}{dx}[fg]\Bigg|_p=f(p)\frac{dg}{dx}\Bigg|_p+g(p)\frac{df}{dx}\Bigg|_p</math> | ||
| − | * | + | * Phonetically ''first times derivative of second plus second times derivative of first'' |
==Example== | ==Example== | ||
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==See also== | ==See also== | ||
* [[Chain rule]] | * [[Chain rule]] | ||
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Latest revision as of 18:45, 28 August 2015
Definition
Given two functions [ilmath]f:\mathbb{R}\rightarrow\mathbb{R} [/ilmath] and [ilmath]g:\mathbb{R}\rightarrow\mathbb{R} [/ilmath] which are differentiable (at [ilmath]p[/ilmath]) the composite function [ilmath]h:\mathbb{R}\rightarrow\mathbb{R} [/ilmath] where [ilmath]h=fg[/ilmath] has derivative:
- [math]\frac{dh}{dx}\Bigg|_p=\frac{d}{dx}[fg]\Bigg|_p=f(p)\frac{dg}{dx}\Bigg|_p+g(p)\frac{df}{dx}\Bigg|_p[/math]
- Phonetically first times derivative of second plus second times derivative of first
Example
- [math]4x^2e^{-x}[/math]
- [math]\frac{d}{dx}\Big[4x^2e^{-x}\Big]=4x^2\frac{d}{dx}\Big[e^{-x}\Big]+e^{-x}\frac{d}{dx}\Big[4x^2\Big][/math]
- [math]=4x^2(-1)e^{-x}+4e^{-x}\frac{d}{dx}\Big[x^2\Big][/math]
- [math]=4e^{-x}\Big(\frac{d}{dx}\Big[x^2\Big]-x^2\Big)[/math]
- [math]=4e^{-x}\big(2x-x^2\big)[/math]
- [math]=4xe^{-x}(2-x)[/math]
- [math]\frac{d}{dx}\Big[4x^2e^{-x}\Big]=4x^2\frac{d}{dx}\Big[e^{-x}\Big]+e^{-x}\frac{d}{dx}\Big[4x^2\Big][/math]
