Question

How do you integrate `f(x)=e^xsinxcosx` using the product rule?

Answer 1

`f'(x) = e^x(sinxcosx + cos^2x - sin^2x)`

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) = e^xsinxcosx` we have;

`{ ("Let "u = e^x, => , (du)/dx = e^x), ("And "v=sinx, =>, (dv)/dx = cosx ), ("And "w = cosx, =>, (dw)/dx = -sinx ) :}`

Applying the product rule we get:

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

`:. f'(x) = (e^x)(sinx)(cosx) + (e^x)(cosx)(cosx) + (e^x)(sinx)(-sinx)`
`" " = e^x(sinxcosx + cos^2x - sin^2x)`