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

1 Answer
Mar 26, 2018

Answer:

#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#