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

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Answer 1 · 2015-06-05T22:59:04

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 · 2015-06-06T14:50:15

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 $cotx$ :

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

$= -2csc^2xcotx-2csc^2xcotx$

$= -4csc^2xcotx$ .

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