Difference between revisions of "Math 22 Chain Rule"

From Math Wiki
Jump to navigation Jump to search
Line 16: Line 16:
 
|<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: &nbsp;
 
!Solution: &nbsp;
 
|-
 
|-
|<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:  

2)

Solution:  

Return to Topics Page

This page were made by Tri Phan