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Derivative of function

by fermio
Tags: derivative, function
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fermio
#1
Jan23-08, 01:46 AM
P: 38
1. The problem statement, all variables and given/known data

[tex]y'=((3x^2+2x+5)^{8x^3+2x^2 +4})'=?[/tex]

2. Relevant equations



3. The attempt at a solution

[tex]((3x^2+2x+5)^{8x^3+2x^2 +4})'=(8x^3+2x^2+4)(3x^2+2x+5)^{8x^3+2x^2 +4-1}(24x^2+4x)(6x+2)[/tex]
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ehrenfest
#2
Jan23-08, 01:55 AM
P: 1,996
The power rule only holds when the exponent is a constant (not a function of x).
Rainbow Child
#3
Jan23-08, 02:02 AM
P: 365
The function [tex]f(x)=g(x)^{h(x)}[/tex] can be written

[tex]f(x)=e^{\ln g(x)^{h(x)}}=e^{h(x)\,\ln g(x)}[/tex]

Now you can take the derivative, i.e.

[tex]f'(x)=e^{h(x)\,\ln g(x)}\left(h(x)\,\ln g(x)\right)'\Rightarrow f'(x)=f(x)\left(h(x)\,\ln g(x)\right)'[/tex]

fermio
#4
Jan23-08, 02:22 AM
P: 38
Derivative of function

[tex]((3x^2+2x+5)^{8x^3+2x^2 +4})'=(3x^2+2x+5)^{8x^3+2x^2 +4}((24x^2+4x)\ln(3x^2+2x+5)+(8x^3+2x^2 +4)\frac{6x+2}{3x^2+2x+5})[/tex]
Rainbow Child
#5
Jan23-08, 02:25 AM
P: 365
You missed a parethensis after [tex](3x^2+2x+5)^{8x^3+2x^2 +4}[/tex], but you are correct
HallsofIvy
#6
Jan23-08, 06:25 AM
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P: 39,683
Quote Quote by Rainbow Child View Post
The function [tex]f(x)=g(x)^{h(x)}[/tex] can be written

[tex]f(x)=e^{\ln g(x)^{h(x)}}=e^{h(x)\,\ln g(x)}[/tex]

Now you can take the derivative, i.e.

[tex]f'(x)=e^{h(x)\,\ln g(x)}\left(h(x)\,\ln g(x)\right)'\Rightarrow f'(x)=f(x)\left(h(x)\,\ln g(x)\right)'[/tex]
Or, much the same thing, write ln(f(x))= h(x)ln(g(x)) and use the product and chain rules: (1/f)f '= h'(x) ln(g(x))+ (h(x)/g(x)) g'(x) so f '= [h'(x) ln(g(x)+(h(x)/g(x))g'(x)]f(x).


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