Question

How do you find the derivative of `cosx/(sinx-2)`?

Answer 1

The answer is: `y'=(2sinx-1)/(sinx-2)^2`

This is because:

`y'=(-sinx*(sinx-2)-cosx*cosx)/(sinx-2)^2=`

`=(-sin^2x+2sinx-cos^2x)/(sinx-2)^2=(2sinx-1)/(sinx-2)^2`

This is because: `sin^2x+cos^2x=1`.