How do you find f' if f(x)=cos^2(3sqrt x )f(x)=cos2(3√x)?

1 Answer
Rohan N.
Mar 26, 2018

f'(x)=(-3 cos(3sqrtx)sin(3sqrtx))/sqrtx

Explanation:

One can use the chain rule break the function into two parts u=3sqrtx or 3x^0.5 and f(u)=(cos(u))^2

Then derive each part x^n becomes nx^(n-1) and cos(x) becomes -sin(x) so

u'=1.5x^-0.5
and f'(u)=2cos(u)*-sin(u) as we have a function inside a function ( )^2 and cos( ) inside it so we have to derive it the outside function and multiply it by the inside function

Then all we have to do now is take replace the u's with 3sqrtx and multiply the very inside derivative u' by f'(u)

f'(x)=(-3 cos(3sqrtx)sin(3sqrtx))/sqrtx

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