Question

How do you differentiate `y = (1 + cos^2x) / (1 - cos^2x)`?

Answer 1

One thing you can do is use some identities so that you can avoid overcomplicating it and having to simplify the results of the quotient rule.

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

so

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

and so

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

`= 2csc^2x - 1`

`(df(x))/(dx) = 2[2cscx*-cscxcotx] = -4csc^2xcotx`

An alternate form of this from Wolfram Alpha is:
`-(16cosx)/(3sinx-sin3x)`

but the accepted answer on Wolfram Alpha is the first one, `-4csc^2xcotx`.

Answer 2

I would rewrite the function:

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

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

`=csc^2x + cot^2x`

Now differentiate using the chain ruloe and the derivatives of `cscx` and `cotx`:

`y' = 2cscx(-cscxcotx) + 2 cotx(-csc^2x)`

`= -2csc^2xcotx-2csc^2xcotx`

`= -4csc^2xcotx`.