# What is the derivative of sqrt(x^2-1)?

##### 1 Answer
Feb 14, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{x}{\sqrt{{x}^{2} - 1}}$

#### Explanation:

Let's equate the function to a variable $y$, so that
$y = \sqrt{{x}^{2} - 1}$

Now, I'll take another variable $t$ and equate it as such,
$t = {x}^{2} - 1$

So that makes the $y$ function as $y = \sqrt{t}$

Now, we are to find the derivative of $y$ with respect to $x$. So that means we are to find $\frac{\mathrm{dy}}{\mathrm{dx}}$

Now, we can use chain rule to simplify our problem as
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{dt}} \cdot \frac{\mathrm{dt}}{\mathrm{dx}}$

That makes it, $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dt}} \left(\sqrt{t}\right) \cdot \frac{d}{\mathrm{dx}} \left({x}^{2} - 1\right)$

Now, we know that for any Real value $n$, $\frac{d}{\mathrm{dx}} \left({x}^{n}\right) = n {x}^{n - 1}$
, and that the derivative of a constant is zero. So
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2 \sqrt{t}} \cdot 2 x$

Now, $t$ was taken as $t = {x}^{2} - 1$, so substituting that back into the equation gives us the answer being searched for (after a few simplifications of course).