How do you differentiate #f(x)=e^(4x)sinx# using the product rule?
1 Answer
Dec 16, 2016
# 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 #