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

2 Answers
Nov 21, 2016

Answer:

#dy/dx = sinx(3cos^2x- 1)#

Explanation:

#y = (1 - cos^2x)cosx = cosx - cos^3x#

We know the derivative of #cosx# is #-sinx#. Letting #y = u^3# and #u = cosx#, we have: #(cos^3x)' = -sinx3u^2 = -sinx3(cosx)^2 =-3cos^2xsinx#

The derivative of the entire expression is:

#dy/dx = -sinx - ( -3cos^2xsinx)#

#dy/dx = 3cos^2xsinx - sinx#

#dy/dx= sinx(3cos^2x- 1)#

Hopefully this helps!

Nov 21, 2016

Answer:

#sin x(3 cos^2 x -1)#

Explanation:

#d/dx (sin^2 x cos x)= sin^2 x d/dx cosx + cos x d/dx sin^2 x#

= # -sin^2 x sinx +cos x (2sinx d/dx sinx) #

= #- sin^3 x +2sin x cos^2 x#

=#sin x (-sin^2 x +2 cos^2x)#

=#sin x (cos^2x -1 + 2 cos^2 x)#

= #sin x(3 cos^2 x -1)#