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

1 Answer
Jun 8, 2016

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