Question

How do you differentiate `g(x) =x^2cscx` using the product rule?

Answer 1

`(dg(x))/dx=2xcscx-x^2cotxcscx`

Explanation:

The product rule states that the derivative of two functions of `x`, let's call them `u` and `v`, is `u'v+uv'`.

First, let's begin by letting `u=x^2` and `v=cscx`, so our example fits with the product rule. Now we can differentiate `u` and `v`:
`u=x^2->u'=2x`
`v=cscx->v'=-cotxcscx`

On to substitutions:
`(dg(x))/dx=(u)'(v)+(u)(v)'`
`(dg(x))/dx=2xcscx-x^2cotxcscx`

This result is fine, but many teachers like the result expressed in sines and cosines. Using some identities, we can do that:
`=(2x)/sinx-(x^2cosx)/sinx*1/sinx`

`=(2x)/sinx-(x^2cosx)/sin^2x`

`=(2xsinx)/sin^2x-(x^2cosx)/sin^2x`

`=(2xsinx-x^2cosx)/sin^2x`