Question

How do you differentiate `sin^2(x) cos (x)`?

Answer 1

`(dy)/(dx)=2sinxcos^2x-sin^3x`

Explanation:

we use the product rule

`y=uv=>`

`(dy)/(dx)=v(du)/(dx)+u(dv)/(dx)`

`y=sin^2xcosx`

`d/(dx)(y)=d/(dx)(sin^2xcosx)`

`(dy)/(dx)=cosxd/(dx)(sin^2x)+sin^2xd/(dx)(cosx)`

`(dy)/(dx)=cosx(2sinxcosx)+sin^2x(-sinx)`

`(dy)/(dx)=2sinxcos^2x-sin^3x`

Answer 2

`f'(x) = -sin x + 3sin x cos^2 x`

Explanation:

An alternative that doesn't use Product Rule and instead relies upon the Chain Rule comes from rewriting the original function prior to differentiating:

`f(x) = sin^2(x) cos(x)`
`f(x) = (1-cos^2 x)*cos(x)`
`f(x) = cos x - cos^3 x`

Thus:

`f'(x) = -sin x - 3*(cos^2 x) *(-sin x)`
`f'(x) = -sin x + 3sin x cos^2 x`

Even though they look different, both answers are the same. ("Peel off" one `sinx cos^2x` term, rewrite `cos^2x` as `1-sin^2x`, and simplify to see.)