Difference between revisions of "Math 22 Chain Rule"
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|<math style="vertical-align: -5px">f(x)=f(x)=\sqrt{x^2+3x-4}=(x^2+3x-4)^{\frac{1}{2}}</math> | |<math style="vertical-align: -5px">f(x)=f(x)=\sqrt{x^2+3x-4}=(x^2+3x-4)^{\frac{1}{2}}</math> | ||
|- | |- | ||
− | |<math>f'(x)=\frac{1}{2}\cdot (x^2+3x-4)^{\frac{1}{2} -1}\frac{d}{dx}[x^2+3x-4]</math> | + | |<math>f'(x)=\frac{1}{2}\cdot (x^2+3x-4)^{(\frac{1}{2} -1)}\frac{d}{dx}[x^2+3x-4]</math> |
|- | |- | ||
− | |<math=(x^2+3x-4)^{\frac{-1}{2}}(2x+3) | + | |<math>=(x^2+3x-4)^{\frac{-1}{2}}(2x+3)</math> |
|} | |} | ||
− | '''2)''' <math>f(x)=(x^2+1)^100</math> | + | '''2)''' <math>f(x)=(x^2+1)^{100}</math> |
{| class = "mw-collapsible mw-collapsed" style = "text-align:left;" | {| class = "mw-collapsible mw-collapsed" style = "text-align:left;" | ||
!Solution: | !Solution: | ||
|- | |- | ||
− | |<math style="vertical-align: -5px">f'(x)=100(x^2+1)^99 \frac{d}{dx}[x^2+1]</math> | + | |<math style="vertical-align: -5px">f'(x)=100(x^2+1)^{99} \frac{d}{dx}[x^2+1]</math> |
|- | |- | ||
− | |<math>=100(x^2+1)^99 (2x)</math> | + | |<math>=100(x^2+1)^{99} (2x)</math> |
|} | |} | ||
Revision as of 06:11, 23 July 2020
The Chain Rule
If is a differentiable function of and is a differentiable function of , then is a differentiable function of and In another word,
Example: Find derivative of
1)
Solution: |
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2)
Solution: |
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This page were made by Tri Phan