How do you differentiate f(x)=e^x*sin^2x using the product rule?

1 Answer
Aug 16, 2017

color(blue)(f'(x) = e^xsinx(sinx + 2cosx)

Explanation:

We're asked to find the derivative

d/(dx) [e^x*sin^2x]

Using the power rule*, which is

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

where

  • u = e^x

  • v = sin^2x:

f'(x) = sin^2xd/(dx)[e^x] + e^xd/(dx)[sin^2x]

The derivative of e^x is e^x:

f'(x) = e^xsin^2x + e^xd/(dx)[sin^2x]

To differentiate the sin^2x term, we'll use the chain rule:

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

where

  • u = sinx

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

f'(x) = e^xsin^2x + e^x*2sinxd/(dx)[sinx]

The derivative of sinx is cosx:

f'(x) = e^xsin^2x + 2e^xsinxcosx

Or

color(blue)(ulbar(|stackrel(" ")(" "f'(x) = e^xsinx(sinx + 2cosx)" ")|)