Question

How do you differentiate `f(x)= xsin3x` using the product rule?

Answer 1

`sin3x+3xcos3x`

Explanation:

The product rule states that:
`d/dxu(x)v(x)=u'(x)v(x)+u(x)v'(x)`

In our example, `u(x)` (the first function) is simply `x`, while `v(x)` (the other function) is `sin3x`:
`x*sin3x`
↑`color(white)(AAA)`↑
`u(x)color(white)()v(x)`

Let's compute the derivatives of `x` and `sin3x` here:
`x'=1`
`(sin3x)'=3cos3x->`(by the chain rule)

Now we plug everything into the chain rule equation:
`d/dxxsin3x=(x)'(sin3x)+(x)(sin3x)'`
`d/dxxsin3x=(1)(sin3x)+(x)(3cos3x)`
`d/dxxsin3x=sin3x+3xcos3x`