What is the derivative of a function raised to another function?

[fermio]

Homework Statement

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

Homework Equations

The Attempt at a Solution

((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)

 

[ehrenfest]

The power rule only holds when the exponent is a constant (not a function of x).

 

[Rainbow Child]

The function f(x)=g(x)^{h(x)} can be written

f(x)=e^{\ln g(x)^{h(x)}}=e^{h(x)\,\ln g(x)}

Now you can take the derivative, i.e.

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)'

 

[fermio]

((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})

 

[Rainbow Child]

You missed a parethensis after (3x^2+2x+5)^{8x^3+2x^2 +4}, but you are correct

 

[HallsofIvy]

The function f(x)=g(x)^{h(x)} can be written

f(x)=e^{\ln g(x)^{h(x)}}=e^{h(x)\,\ln g(x)}

Now you can take the derivative, i.e.

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)'

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).