Question

What is the derivative of sin x / cos ^2 x - 1 ?

I come to the answer `(-cos x sin^2 x)/((cos^2x -1)^2)`

But it should be :
`(-cos x )/((cos^2x -1))`

I know `sin^2x = 1 - cos^2x`

But its not the same as `cos^2x -1`

So I do something wrong.

My first step was :

`((-2 cos x sinx sinx ) - (cos x (cos^2x -1)))/ (cos^2x -1)^2`

Simplify
`((-2cosx sin^2x) - cos x (cos^2x-1))/((cos^2x -1)^2)`

Gives:
`(-cosx((cos^2x-1)+2 sin^2x))/((cos^2x -1)^2)`

Gives:
`(-cosx(1+sin^2x-1))/((cos^2x -1)^2)`

Gives:
`(-cosxsin^2x)/((cos^2x -1)^2)`

Answer 1

You're almost there!

What you have to realize is that `cos^2x-1=-(1-cos^2x)`. Try distributing the minus sign and see what you get.

Once you understand that, it's just one step further to say that `cos^2x-1=-sin^2x`. You can also get this equation straight away by moving terms around in `sin^2x+cos^2x=1` (move the `1` and `sin^2x` to their opposite sides).

Another error you have is that you reversed the order of the terms in the numerator when you did the quotient rule, remember, `d/dx(u/v)=((du)/dx*v-u*(dv)/dx)/v^2`.

This makes what you got off by a factor of `-1`, so you should've gotten just `(cosxsin^2x)/(cos^2x-1)^2` as your answer (I took away the minus sign).

From here, use the identity `cos^2x-1=-sin^2x`:

`(cosxsin^2x)/(cos^2x-1)^2=(cosxsin^2x)/((cos^2x-1)(cos^2x-1))`

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

`=(-cosx)/(cos^2x-1)`

As you wished to show.

But, honestly, I think the best way to express this is by rewriting the remaining `cos^2x-1`:

`=(-cosx)/(-sin^2x)`

`=cosx/sinx(1/sinx)`

`=cotxcscx`

If you know your trigonometric derivatives, you might recognize that this is off the derivative of `cscx` by a factor of `-1`. This isn't surprising, though, when we consider the original function:

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

So the derivative is exactly as we'd expect.