# How do you differentiate #f(x)=e^(4x)sinx# using the product rule?

##### 1 Answer

Dec 16, 2016

#### Answer:

# f'(x) = e^(4x)cosx + 4e^(4x)sinx #

#### Explanation:

If you are studying maths, then you should learn the Product Rule for Differentiation, and practice how to use it:

# d/dx(uv)=u(dv)/dx+(du)/dxv # , or,# (uv)' = (du)v + u(dv) #

I was taught to remember the rule in words; "*The first times the derivative of the second plus the derivative of the first times the second* ".

This can be extended to three products:

# d/dx(uvw)=uv(dw)/dx+u(dv)/dxw + (du)/dxvw#

So with

Applying the product rule we get:

# " " d/dx(uv)=u(dv)/dx + (du)/dxv #

# d/dx(e^(4x)sinx)=(e^(4x))(cosx) + (4e^(4x))(sinx) #

# d/dx(e^(4x)sinx)=e^(4x)cosx + 4e^(4x)sinx #