Question

What is the derivative of `y = (cosx-sinx)/(cosx+sinx)`?

Answer 1

`y=frac{cosx-sinx}{cosx+sinx}`

Use the quotient rule:
`dy/dx = frac{(cosx+sinx)(-sinx-cosx)-(cosx-sinx)(-sinx+cosx)}{(cosx + sinx)(cosx+sinx)}`

Distribute the terms to simplify
`= frac{-sinxcosx-cos^2x-sin^2x-sinxcosx -(-sinxcosx+cos^2x+sin^2x-sinxcosx)}{cos^2x+2sinxcosx+sin^2x}`

`= frac{-2sinxcosx-1-(-2sinxcosx+1)}{2sinxcosx+1}`

`= frac{2(2sinxcosx-1)}{2sinxcosx+1}`

Multiply by the conjugate of the denominator over itself:
`= frac{2(2sinxcosx-1)}{2sinxcosx+1}*frac{2sinxcosx-1}{2sinxcosx-1}`

`= frac{2(2sinxcosx-1)^2}{4sin^2xcos^2x-1}`

Answer 2

`dy/dx = 2sec2x(tan2x - sec 2x)`

Explanation:

We have:

`y = (cosx-sinx)/(cosx+sinx)`

We can write:

`y = (cosx-sinx)/(cosx+sinx) * (cosx-sinx)/(cosx-sinx)`

`\ \ = (cos^2x-2sinxcosx+sin^2x)/(cos^2x-sin^2x)`

`\ \ = (1-sin2x)/(cos2x)`

`\ \ = sec2x-tan2x`

Then differentiating wrt `x`:

`dy/dx = 2sec2xtan2x - 2sec^2 2x`
`\ \ \ \ \ = 2sec2x(tan2x - sec 2x)`