How do you differentiate f(x)=e^cos(-2lnx) using the chain rule?

Aug 25, 2016

$f ' \left(x\right) = \frac{2 \sin \left(- 2 \ln \left(x\right)\right) \times {e}^{\cos \left(- 2 \ln \left(x\right)\right)}}{x}$

Explanation:

Let's first consider a few derivative techniques and procedures:

If $F \left(x\right) = f \left(g \left(n x\right)\right)$, then the derivative of the function may be written as: $F ' \left(x\right) = f ' \left(g \left(x\right)\right) \times g ' \left(x\right) \times n$

*The derivative of ${e}^{k x}$ if $k {e}^{k x}$ *

The derivative of $\cos \left(n x\right)$ is $- n \sin \left(x\right)$

The derivative of $\ln \left(a x\right)$ is $\frac{1}{a x} \times a$

We can structure your question in a similar manner:

If we consider each "section" of the function separately, we can state that:

f'(x)=(d{e^(cos(-2ln(x)))}}/(d{cos(-2ln(x))}

xx (d{cos(-2ln(x)))}/(d{(-2ln(x))}

xx (d{-2ln(x))}/(d{x}

Applying our rules from above, we can derive such that:

$f ' \left(x\right) = \cos \left(- 2 \ln \left(x\right)\right) \times {e}^{\cos \left(- 2 \ln \left(x\right)\right)} \times - \sin \left(- 2 \ln \left(x\right)\right) \times - 2 \times \frac{1}{x}$

Simplifying this we get:

$f ' \left(x\right) = - 2 \times - \sin \left(- 2 \ln \left(x\right)\right) \times {e}^{\cos \left(- 2 \ln \left(x\right)\right)} \times \frac{1}{x}$

Simplifying once more we get:

$f ' \left(x\right) = \frac{2 \sin \left(- 2 \ln \left(x\right)\right) \times {e}^{\cos \left(- 2 \ln \left(x\right)\right)}}{x}$