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

1 Answer
Aug 25, 2016

f'(x)= (2sin(-2ln(x))xxe^(cos(-2ln(x))))/x

Explanation:

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

If F(x)=f(g(nx)), then the derivative of the function may be written as: F'(x)=f'(g(x))xxg'(x)xxn

*The derivative of e^(kx) if ke^(kx) *

The derivative of cos(nx) is -nsin(x)

The derivative of ln(ax) is 1/(ax) xx 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'(x)= cos(-2ln(x)) xx e^(cos(-2ln(x)))xx -sin(-2ln(x)) xx -2xx 1/x

Simplifying this we get:

f'(x)= -2xx-sin(-2ln(x))xxe^(cos(-2ln(x)))xx1/x

Simplifying once more we get:

f'(x)= (2sin(-2ln(x))xxe^(cos(-2ln(x))))/x