What is the derivative of #cos^2(x^3)#?

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Answer 1 · 2016-01-22T13:00:46

#f'(x) = -6 x^2 cos (x^3) sin (x^3)#

#f(x) = cos^2(x^3)#

Let's break your function down as a chain of functions:

#f(x) = [color(blue)(cos (x^3))]^2 = color(blue)(u)^2#

where

#u = cos(color(violet)(x^3)) = cos(color(violet)(v))#

where

#v = x^3#

Thus, the derivative of #f(x)# is:

#f'(x) = [ u^2 ]' * u' = [u^2]' * [cos v]' * v'#

Now, let's compute those three derivatives:

#[u^2]' = 2u = 2 cos x^3#

#[cos v]' = - sin v = - sin x^3#

#[ v]' = [x^3]' = 3x^2#

Thus, you can compute your derivative as follows:

#f'(x) = [u^2]' * [cos v]' * v'#

# = 2 cos x^3 * (- sin x^3) * 3x^2#

# = -6 x^2 cos (x^3) sin (x^3)#