Question

How do you find the derivative of `(sin x + 2x) / (cos x - 2)`?

Answer 1

`f'(x) = (2xsinx-3)/(cosx-2)^2`

Explanation:

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

This can be differentiate using quotient rule

`f'(x) = [(cosx-2)*d/(dx)(sinx +2x)- (sinx+2x)*d/(dx)(cosx-2)]/(cosx-2)^2`

`=>f'(x) = [(cos x-2)(cos x+2)-(sinx+2x)(-sinx)]/(cosx-2)^2`

`=>f'(x) = (cos^2x -4 + sin^2x +2x sinx)/(cosx-2)^2`

`=>f'(x) = (color(red)(cos^2x +sin^2x)-4 +2x sinx)/(cosx-2)^2`

`=>f'(x) = (color(red)(1)-4+2x sinx)/(cosx-2)^2`

`f'(x) = (2xsinx-3)/(cosx-2)^2`

*`sin^2x + cos^2x = 1`