How do you find the derivative of the square root of (sinxcosx) step by step

thanks!

First, to initiate the chain rule, do a u-substitution where u=sin(x)cos(x) and d(sqrt(u))/du=1/(2sqrt(u)):

This gives us d/dx [(sin(x)cos(x))/(2sqrt(sin(x)cos(x))].

Next use the product rule where u=cos(x) and v=sin(x):

This gives us (-sin(x)sin(x)+(cos(x))^2)/(2sqrt(sin(x)… and would be an acceptable answer to most teachers.

This can further be simplified to an alternate form (if you want) to [cos(2x)]/[(2sqrt(sin(x)cos(x))].

d/dx (sinxcosx)^(1/2) = √(sinxcosx)

chain rule and product rule:
(1/2)[(sinxcosx)]^-(1/2) * [sinx*(-sinx) + cosx*(cosx)] =
(1/2)[(sinxcosx)]^-(1/2) * [-sin^2x + cos^2x] =
[-sin^2x + cos^2x] / 2√(sinxcosx)

Chain Rule:

Sqrt(sinxcosx)

1/2(sinxcosx)^-1/2 *(sinx*-sinx + (cosx)^2) <---- product rule for the inside


1/2(1/sqrt(sinxcosx))*(-sin^2x + (cosx)^2)