# How do you differentiate f(x)=cot(sqrt(x^2-1))  using the chain rule?

##### 1 Answer
Dec 22, 2015

The explanation is given below.

#### Explanation:

$f \left(x\right) = \cot \left(\sqrt{{x}^{2} - 1}\right)$

Let us first write the same as the following.

$y = \cot \left(u\right)$ and $u = \sqrt{v}$ and $v = {x}^{2} - 1$
The above steps are like breaking the chain into manageable links.

Chain rule would work like this:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dv}} \cdot \frac{\mathrm{dv}}{\mathrm{dx}}$ Chain rule

$y = \cot \left(u\right)$
$\frac{\mathrm{dy}}{\mathrm{du}} = - {\sec}^{2} \left(u\right)$

$u = \sqrt{v}$
$\frac{\mathrm{du}}{\mathrm{dv}} = \frac{1}{2 \sqrt{v}}$

$v = {x}^{2} - 1$
$\frac{\mathrm{dv}}{\mathrm{dx}} = 2 x$

$f ' \left(x\right) = \frac{\mathrm{dy}}{\mathrm{dx}} = - {\sec}^{2} \left(u\right) \cdot \frac{1}{2 \sqrt{v}} \cdot 2 x$
Simplifying

$f ' \left(x\right) = \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- x \cdot {\sec}^{2} \left(\sqrt{v}\right)}{\sqrt{v}}$
$f ' \left(x\right) = \frac{\mathrm{dy}}{\mathrm{dx}} = \left(- x \cdot {\sec}^{2} \frac{\sqrt{{x}^{2} - 1}}{\sqrt{{x}^{2} - 1}}\right)$

$f ' \left(x\right) = \left(- x \cdot {\sec}^{2} \frac{\sqrt{{x}^{2} - 1}}{\sqrt{{x}^{2} - 1}}\right)$ answer