Derivative of a composite function?

potmobius

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

Find the derivative of the function: cos(x)^(cos(cos(x)))

2. Relevant equations

The chain rule

3. The attempt at a solution

I know how the chain rule works, and I've done many problems with composite functions. However, I just don't know where to start with this one. I'm lost and confused :(

Elucidus

Correct me if I am wrong. It seems that you are trying to find

[tex]\frac{d}{dx}\;(\cos x)^{\cos(\cos x)}[/tex]

Any time you need to differentiate an expression that involves a variable base and exponent (as we have here) you need to use logarithms and implicit differentiation.

i.e.

[tex]y=f(x)^{g(x)}[/tex]

[tex]\ln y = \ln \left (f(x)^{g(x)} \right)[/tex]

[tex]\ln y = g(x) \cdot \ln (f(x))[/tex]

[tex]\frac{d}{dx} \ln y = \frac{d}{dx} \left[ g(x) \cdot \ln (f(x)) \right] [/tex]

[tex]\frac{y'}{y} = \frac{g(x) \cdot f'(x)}{f(x)} + g'(x) \cdot \ln (f(x)) [/tex]

[tex] y' = y \cdot \left[ \frac{g(x) \cdot f'(x)}{f(x)} + g'(x) \cdot \ln (f(x)) \right] [/tex]

[tex] y' = \left( f(x)^{g(x)} \right) \cdot \left[ \frac{g(x) \cdot f'(x)}{f(x)} + g'(x) \cdot \ln (f(x)) \right] [/tex]

Hopefully this will get you going in the right direction.

--Elucidus

Dick

I'm glad to hear you know how the chain rule works. Now prove it. cos(x) is exp(log(cos(x)). Does that help? Now use properties of exponents and the chain rule. You'll need some product rule as well.

potmobius

Yes, that's exactly what I meant, and this clears it up! Thanks :)