# Derivative of function

1. Jan 23, 2008

### fermio

1. The problem statement, all variables and given/known data

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

2. Relevant equations

3. The attempt at a solution

$$((3x^2+2x+5)^{8x^3+2x^2 +4})'=(8x^3+2x^2+4)(3x^2+2x+5)^{8x^3+2x^2 +4-1}(24x^2+4x)(6x+2)$$

2. Jan 23, 2008

### ehrenfest

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

Last edited: Jan 23, 2008
3. Jan 23, 2008

### 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.

$$f'(x)=e^{h(x)\,\ln g(x)}\left(h(x)\,\ln g(x)\right)'\Rightarrow f'(x)=f(x)\left(h(x)\,\ln g(x)\right)'$$

4. Jan 23, 2008

### fermio

$$((3x^2+2x+5)^{8x^3+2x^2 +4})'=(3x^2+2x+5)^{8x^3+2x^2 +4}((24x^2+4x)\ln(3x^2+2x+5)+(8x^3+2x^2 +4)\frac{6x+2}{3x^2+2x+5})$$

5. Jan 23, 2008

### Rainbow Child

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

6. Jan 23, 2008

### HallsofIvy

Staff Emeritus
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).