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

2 Answers
Oct 2, 2017

(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

Oct 2, 2017

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.)