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

Jan 19, 2016

$f ' \left(x\right) = \frac{- 4 \cos x}{{\sin}^{3} x}$

#### 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

$f \left(x\right) = \frac{1 + {\cos}^{2} x}{\sin} ^ 2 x$

The quotient rule states that for $f \left(x\right) = \frac{g \left(x\right)}{h \left(x\right)}$, the derivative is

$f ' \left(x\right) = \frac{g ' \left(x\right) h \left(x\right) - h ' \left(x\right) g \left(x\right)}{{h}^{2} \left(x\right)}$

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

$g \left(x\right) = 1 + {\cos}^{2} x \textcolor{w h i t e}{\times} \implies \textcolor{w h i t e}{\times} g ' \left(x\right) = - 2 \cos x \sin x$

$h \left(x\right) = {\sin}^{2} \textcolor{w h i t e}{\times} \implies \textcolor{w h i t e}{\times} h ' \left(x\right) = 2 \sin x \cos x$

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

$f ' \left(x\right) = \frac{- 2 \cos x \sin x \cdot {\sin}^{2} x - 2 \sin x \cos x \left(1 + {\cos}^{2} x\right)}{{\sin}^{4} x}$

$\textcolor{w h i t e}{\times \times} = \frac{\sin x \cdot \left(- 2 \cos x {\sin}^{2} x - 2 \cos x \left(1 + {\cos}^{2} x\right)\right)}{\sin x \cdot {\sin}^{3} x}$

...cancel $\sin x$ ...

$\textcolor{w h i t e}{\times \times} = \frac{\cancel{\sin x} \cdot \left(- 2 \cos x {\sin}^{2} x - 2 \cos x \left(1 + {\cos}^{2} x\right)\right)}{\cancel{\sin x} \cdot {\sin}^{3} x}$

$\textcolor{w h i t e}{\times \times} = \frac{\textcolor{b l u e}{- 2 \cos x {\sin}^{2} x} - 2 \cos x \textcolor{b l u e}{- 2 \cos x \cdot {\cos}^{2} x}}{{\sin}^{3} x}$

$\textcolor{w h i t e}{\times \times} = \frac{\textcolor{b l u e}{- 2 \cos x \left({\sin}^{2} x + {\cos}^{2} x\right)} - 2 \cos x}{{\sin}^{3} x}$

... use ${\sin}^{2} x + {\cos}^{2} x = 1$ once more ...

$\textcolor{w h i t e}{\times \times} = \frac{- 2 \cos x - 2 \cos x}{{\sin}^{3} x}$

$\textcolor{w h i t e}{\times \times} = \frac{- 4 \cos x}{{\sin}^{3} x}$

Jan 19, 2016

$y ' = - 4 \cos \frac{x}{\sin} ^ 3 \left(x\right)$

#### Explanation:

Alternatively, to more simplify the computation:

${\sin}^{2} \left(x\right) + {\cos}^{2} \left(x\right) = 1$

therefore

${\cos}^{2} \left(x\right) = 1 - {\sin}^{2} \left(x\right)$

$y = \frac{1 + {\cos}^{2} \left(x\right)}{1 - {\cos}^{2} \left(x\right)} = \frac{1 + 1 - {\sin}^{2} x}{\textcolor{red}{\cancel{1}} \textcolor{red}{\cancel{- 1}} + {\sin}^{2} \left(x\right)} = \frac{2 - {\sin}^{2} \left(x\right)}{\sin} ^ 2 \left(x\right) =$
$= \frac{2}{\sin} ^ 2 \left(x\right) - 1 = 2 {\left[\sin \left(x\right)\right]}^{- 2} - 1$

Applying the Power Rule and Chain Rule

$y ' = 2 \cdot \left(- 2 \cdot {\left[\sin \left(x\right)\right]}^{- 2 - 1} \cdot \cos \left(x\right)\right) - 0 =$

$= - 4 \cdot {\left[\sin \left(x\right)\right]}^{- 3} \cos x = - 4 \cos \frac{x}{\sin} ^ 3 \left(x\right)$

Alternatively

$y = \frac{2}{\sin} ^ 2 \left(x\right) - 1$

Using the Quotient Rule

$y ' = \frac{0 \cdot {\sin}^{2} \left(x\right) - 2 \sin \left(x\right) \cos \left(x\right) \cdot 2}{{\sin}^{2} \left(x\right)} ^ 2 + 0 =$

$= - 4 \textcolor{red}{\cancel{\sin x}} \cos \frac{x}{\sin x} ^ \left({\textcolor{red}{\cancel{4}}}^{3}\right) =$

$- 4 \cos \frac{x}{\sin} ^ 3 \left(x\right)$