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 # f(x) = x^2sinxtanx # we have;

# { ("Let "u = e^(4x), => , (du)/dx = 4e^(4x)'), ("And "v = sinx, =>, (dv)/dx = cosx' ) :}#

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 #