Question

How do you differentiate `f(x)=-2xsin^2(x)` using the product rule?

Answer 1

`f'(x)=-2sinx(sinx+2xcosx)`

Explanation:

Product rule: for a function

`f(x)=g(x)h(x)`,

`f'(x)=g'(x)h(x)+h'(x)g(x)`

So,

`g(x)=-2x`
`h(x)=sin^2x`

Find the derivative of each.

`g'(x)=-2`

Use the chain rule to find `h'(x)`, recalling that `d/dx u^2=2u*(du)/dx`.

Thus,

`h'(x)=2sinxd/dxsinx=2sinxcosx`

Put this all together:

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

Simplify.

`f'(x)=-2sinx(sinx+2xcosx)`

Answer 2

I remember the product rule by remembering the phrase:

"first, d-second, plus second, d-first"

or:

`color(green)(d/(dx)[g(x)h(x)] = g(x)(dh(x))/(dx) + h(x)(dg(x))/(dx))`

The derivative here requires that you use the chain rule on `sin^2(x)`, so let's check that out first:

`color(green)(d/(dx)[f(u(x))] = (df(u(x)))/cancel(du) * cancel(du)/(dx) = (df(u(x)))/(dx))`

If we let `u(x) = sinx`, then we should get `2u * (du)/(dx)` like so:

`d/(dx)[(sinx)^2] = 2sinx * d/(dx)[sinx] = 2sinxcosx`

So, the final result is:

`d/(dx)[-2xsin^2x] = (-2x)(2sinxcosx) + (sin^2x)(-2)`

`= color(blue)(-4xsinxcosx - 2sin^2x)`