How do you find the derivative of #(sin^2x)(cos^3x)#?

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Answer 1 · 2017-07-13T15:08:00

#d/(dx)[(cos^3x)(sin^2x)] = color(blue)(2cos^4xsinx - 3cos^2xsin^3x#

We're asked to find the derivative

#d/(dx) [(cos^3x)(sin^2x)]#

The first step we could do is use the product rule, which is

#d/(dx) [uv] = v(du)/(dx) + u(dv)/(dx)#

where in this case

#= cos^3x(d/(dx)[sin^2x]) + sin^2x(d/(dx)[cos^3x))#

We can differentiate the #sin^2x# term via the chain rule, which would look like

#d/(dx)[sin^2x] = (du^2)/(du)(du)/(dx)#

where

#d/(du) [u^2] = 2u# (power rule):

#= cos^3x(2d/(dx)[sinx]sinx) + sin^2x(d/(dx)[cos^3x))#

The derivative of #sinx# is #cosx#:

#= cos^3x(2cosxsinx) + sin^2x(d/(dx)[cos^3x))#

Simplifying:

#= 2cos^4xsinx + d/(dx)[cos^3x]sin^2x#

We now use the chain rule again for differentiating the #cos^3x# term:

#d/(dx)[cos^3x] = (du^3)/(du)(du)/(dx)#

where

#= 2cos^4xsinx + 3cos^2xd/(dx)[cosx]sin^2x#

The derivative of #cosx# is #-sinx#:

#= 2cos^4xsinx - 3cos^2xsinxsin^2x#

Simpligying gives

#= color(blue)(2cos^4xsinx - 3cos^2xsin^3x#