Question

How do you differentiate `f(x)=(x^2-2x+8)/sinx-1/x` using the quotient rule?

Answer 1

`f'(x) = (2x-2)/sinx - ((x^2-2x+8)cosx)/sin^2x - x^-2`

Explanation:

The quotient rule is `f'(x) = (v(du)/(dx) - u(dv)/(dx))/v^2`

First lets rearrange the second term in f(x) to make it easier to differentiate.

`f(x) = (x^2 - 2x + 8)/sinx - x^-1`

Next let's differentiate the first term using the quotient rule.

`u = x^2 - 2x + 8`
`v = sinx`

`(du)/dx = 2x - 2`

`(dv)/dx = cosx`

`f'(x) = ((2x-2)sinx - (x^2-2x + 8)cosx)/(sinx)^2`

We can simplify this a bit by splitting up the fraction and cancelling the sin on the top and the bottom.

`f'(x) = (2x-2)/sinx - ((x^2-2x+8)cosx)/sin^2x`

Now we just put that back in and differentiate `x^-1`

`f'(x) = (2x-2)/sinx - ((x^2-2x+8)cosx)/sin^2x + x^-2`