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derivative of function |
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| Jan23-08, 01:46 AM | #1 |
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derivative of function
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] |
| Jan23-08, 01:55 AM | #2 |
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The power rule only holds when the exponent is a constant (not a function of x).
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| Jan23-08, 02:02 AM | #3 |
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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] |
| Jan23-08, 02:22 AM | #4 |
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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]
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| Jan23-08, 02:25 AM | #5 |
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You missed a parethensis after [tex](3x^2+2x+5)^{8x^3+2x^2 +4}[/tex], but you are correct
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| Jan23-08, 06:25 AM | #6 |
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