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

1 Answer
Aug 16, 2017

Answer:

#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)" ")|)#