Question

How do you differentiate `cosx/(sinx-2)`?

Answer 1

`f(x)=(cosx)/(sinx-2)`

Use quotient rule:
`f'(x)=[(sinx-2)(-sinx)-(cosx)(cosx)]/[(sinx-2)^2]`

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

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

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