Math 22 Chain Rule - Revision history

Tphan046: /* The Chain Rule */

2020-07-23T14:16:38Z

The Chain Rule

2020-07-23T14:11:58Z

The Chain Rule

2020-07-23T14:10:36Z

The Chain Rule

2020-07-23T13:58:27Z

2020-07-23T13:53:30Z

Created page with " '''Return to Topics Page''' '''This page were made by Tri Phan'''"

New page

[[Math_22| '''Return to Topics Page''']]

'''This page were made by [[Contributors|Tri Phan]]'''

@@ 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>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;"
 !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>
 |}

@@ Line 29: / Line 29: @@
 |<math>=100(x^2+1)^{99} (2x)</math>
 |}
 [[Math_22| '''Return to Topics Page''']]
 '''This page were made by [[Contributors|Tri Phan]]'''