Question

How do you find the derivative of `y = (1 + cos^2x) / (1 - cos^2x)`?

Answer 1

`f'(x) = (- 4 cos x) / (sin^3 x)`

Explanation:

First of all, use the fact that `sin^2 x + cos^2 x = 1`.

Thus, `1 - cos^2 x = sin^2 x` holds.

So, you have the function

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

The quotient rule states that for `f(x) = (g(x)) / (h(x))`, the derivative is

`f'(x) = (g'(x) h(x) - h'(x) g(x)) / (h^2(x))`

In your case, using the chain rule along the way, you get:

`g(x) = 1 + cos^2 x color(white)(xx) => color(white)(xx) g'(x) = - 2 cos x sin x`

`h(x) = sin^2 color(white)(xx) => color(white)(xx) h'(x) = 2 sin x cos x`

Now you need to use the quotient rule and you will get:

`f'(x) = (- 2 cos x sin x * sin^2 x - 2 sin x cos x (1 + cos^2 x) ) / (sin^4 x)`

`color(white)(xxxx) =(sin x * (- 2 cos x sin^2 x - 2 cos x (1 + cos^2 x) )) / (sin x * sin^3 x)`

...cancel `sin x` ...

`color(white)(xxxx) =(cancel(sin x) * (- 2 cos x sin^2 x - 2 cos x (1 + cos^2 x) )) / (cancel(sin x) * sin^3 x)`

`color(white)(xxxx) = (color(blue)(- 2 cos x sin^2 x) - 2 cos x color(blue)( - 2 cos x * cos^2 x)) / (sin^3 x)`

`color(white)(xxxx) = (color(blue)(- 2 cos x (sin^2 x + cos^2 x)) - 2 cos x ) / (sin^3 x)`

... use `sin ^2 x + cos^2 x = 1` once more ...

`color(white)(xxxx) = (- 2 cos x- 2 cos x ) / (sin^3 x)`

`color(white)(xxxx) = (- 4 cos x) / (sin^3 x)`

Answer 2

`y'=-4cos(x)/sin^3(x)`

Explanation:

Alternatively, to more simplify the computation:

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

therefore

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

`y=(1+cos^2(x))/(1-cos^2(x))=(1+1-sin^2x)/(color(red)cancel(1)color(red)cancel(-1)+sin^2(x))=(2-sin^2(x))/sin^2(x)=`
`=2/sin^2(x)-1=2[sin(x)]^(-2)-1`

Applying the Power Rule and Chain Rule

`y'=2*(-2*[sin(x)]^(-2-1)*cos(x))-0=`

`=-4*[sin(x)]^(-3)cosx=-4cosx/sin^3(x)`

Alternatively

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

Using the Quotient Rule

`y'=(0*sin^2(x)-2sin(x)cos(x)*2)/[sin^2(x)]^2+0=`

`=-4color(red)cancel(sinx)cosx/[sinx]^(color(red)cancel(4)^3)=`

`-4cos(x)/sin^3(x)`