How do you find the derivative of #f(x)=sin^2(cos^2(x))#?

1 Answer
Oct 21, 2016

#f'(x)=-4sin(cos^2(x))cos(cos^2(x))sin(x)cos(x)#

Explanation:

We have many instances of the chain rule here. The trick is to think of a function like an onion--the function contains other functions.

The first function, the most pressing one, is the squared function.

#f(x)=sin^2(cos^2(x))# is a squared function in the form #(g(x))^2#. The chain rule tells us that the derivative of this function is #2g(x)*g'(x)#, through the power rule with the chain rule.

Thus, #f'(x)=2sin(cos^2(x))*[d/dxsin(cos^2(x))]#. To find this derivative, the most pressing function is the sine function. We see that this is in the form #sin(g(x))#, and since the derivative of sine is cosine, we see that this function composition's derivative is #cos(g(x))*g'(x)#.

Therefore, #f'(x)=2sin(cos^2(x)) * [cos(cos^2(x)) * {d/dxcos^2(x)}]#.

To find the derivative now, we again have a squared function. Using the rule outlined above, we see that #f'(x)=2sin(cos^2(x)) * cos(cos^2(x)) * {2cos(x) * d/dxcos(x)}#.

The derivative of cosine is the opposite of sine, so combining all these, we see that #f'(x)=2sin(cos^2(x)) * cos(cos^2(x)) * 2cos(x) * (-sin(x))#.

Combined, this looks like #f'(x)=-4sin(cos^2(x))cos(cos^2(x))sin(x)cos(x)#.