Question

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

Answer 1

`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`