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

1 Answer
Aug 25, 2016

Answer:

#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#