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

2 Answers

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)

Jan 19, 2016

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)