# Find the derivative?

## y= x sqrt(x^2-4

Apr 25, 2018

See the answer for the process on arriving to:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 \left({x}^{2} - 2\right)}{\sqrt{{x}^{2} - 4}}$

#### Explanation:

To find the derivative of

$y = x \sqrt{{x}^{2} - 4}$

we will (first) need to use the product rule. Recall that the product rule states that the derivative of the product of functions $f$ and $g$ is given by ${\left(f g\right)}^{'} = {f}^{'} g + f {g}^{'}$.

The two functions being multiplied here are $x$ and $\sqrt{{x}^{2} - 4}$, so we see that the derivative of $y$ is given by

$\frac{\mathrm{dy}}{\mathrm{dx}} = \left(\frac{d}{\mathrm{dx}} x\right) \sqrt{{x}^{2} - 4} + x \left(\frac{d}{\mathrm{dx}} \sqrt{{x}^{2} - 4}\right)$

Note that $\frac{d}{\mathrm{dx}} x = 1$. In order to find $\frac{d}{\mathrm{dx}} \sqrt{{x}^{2} - 4}$, we will need the chain rule.

First, recall that $\sqrt{{x}^{2} - 4} = {\left({x}^{2} - 4\right)}^{\frac{1}{2}}$. We differentiate this as we do ${x}^{\frac{1}{2}}$, with the power rule, bearing in mind that instead of just $x$ we're working with the more complex function ${x}^{2} - 4$. After applying the power rule, we use the chain rule, and multiply by the derivative of the inner function ${x}^{2} - 4$.

In all, we see that

$\frac{d}{\mathrm{dx}} {\left({x}^{2} - 4\right)}^{\frac{1}{2}} = \frac{1}{2} {\left({x}^{2} - 4\right)}^{- \frac{1}{2}} \left(\frac{d}{\mathrm{dx}} \left({x}^{2} - 4\right)\right)$

The derivative of the inner function is $2 x$, so we see that

$\frac{d}{\mathrm{dx}} {\left({x}^{2} - 4\right)}^{\frac{1}{2}} = \frac{1}{2} {\left({x}^{2} - 4\right)}^{- \frac{1}{2}} \left(2 x\right) = \frac{x}{\sqrt{{x}^{2} - 4}}$

Returning to the whole function, substitute the two derivatives we've found in:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \left(1\right) \sqrt{{x}^{2} - 4} + x \left(\frac{x}{\sqrt{{x}^{2} - 4}}\right)$

And simplifying:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \sqrt{{x}^{2} - 4} + {x}^{2} / \sqrt{{x}^{2} - 4}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left({x}^{2} - 4\right) + {x}^{2}}{\sqrt{{x}^{2} - 4}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 \left({x}^{2} - 2\right)}{\sqrt{{x}^{2} - 4}}$

Apr 25, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 {x}^{2} - 4}{{x}^{2} - 4} ^ \left(\frac{1}{2}\right)$

#### Explanation:

We're attempting to find the derivative of the product of two things, so the Product Rule will help here.

First, I'll rewrite our equation in terms of functions. Thus, we have:

$y = f \left(x\right) g \left(x\right)$ where

$f \left(x\right) = x \implies \textcolor{b l u e}{f ' \left(x\right) = 1}$

$g \left(x\right) = \sqrt{{x}^{2} - 4} \implies \textcolor{\lim e}{g ' \left(x\right) = \frac{x}{\sqrt{{x}^{2} - 4}}}$

NOTE: $\textcolor{\lim e}{g ' \left(x\right)}$ found via Chain Rule- Inside function (${x}^{2} - 4$), outside function (${x}^{\frac{1}{2}}$)

Product Rule:

$f \left(x\right) g ' \left(x\right) + f ' \left(x\right) g \left(x\right)$

Since we know both functions and their derivatives, we can plug in now. We get:

$\frac{\mathrm{dy}}{\mathrm{dx}} = x \cdot \textcolor{\lim e}{\frac{x}{\sqrt{{x}^{2} - 4}}} + \textcolor{b l u e}{1} \cdot \sqrt{{x}^{2} - 4}$

$= \frac{{x}^{2}}{\sqrt{{x}^{2} - 4}} + \sqrt{{x}^{2} - 4}$

$= \frac{{x}^{2}}{{\left({x}^{2} - 4\right)}^{\frac{1}{2}}} + \left(\frac{{\left({x}^{2} - 4\right)}^{\frac{1}{2}}}{1} \cdot \textcolor{red}{\frac{{\left({x}^{2} - 4\right)}^{\frac{1}{2}}}{{\left({x}^{2} - 4\right)}^{\frac{1}{2}}}}\right)$

NOTE: We multiplied by the red expression to find a common denominator

$= \frac{{x}^{2}}{{\left({x}^{2} - 4\right)}^{\frac{1}{2}}} + \left(\frac{{\left({x}^{2} - 4\right)}^{\frac{1}{2} + \frac{1}{2}}}{{x}^{2} - 4} ^ \left(\frac{1}{2}\right)\right)$

$= \frac{{x}^{2} + {x}^{2} - 4}{{x}^{2} - 4} ^ \left(\frac{1}{2}\right)$

$\textcolor{p u r p \le}{\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 {x}^{2} - 4}{{x}^{2} - 4} ^ \left(\frac{1}{2}\right)}$

After using the Product Rule and a good deal of algebraic manipulation, we were able to find the derivative of $y$.

Hope this helps!