# How do I find f'(3) given that f(x)=sqrt(3x)/(x^2-4)?

May 29, 2017

$f ' \left(3\right) = - \frac{31}{50}$

#### Explanation:

Let $f \left(x\right) = \frac{\sqrt{3 x}}{{x}^{2} - 4}$

$\frac{d}{\mathrm{dx}} \left(\frac{p \left(x\right)}{q \left(x\right)}\right) = \frac{q \left(x\right) p ' \left(x\right) - p \left(x\right) q ' \left(x\right)}{q {\left(x\right)}^{2}}$

Let $p \left(x\right) = \sqrt{3 x}$, then $p ' \left(x\right) = \frac{\sqrt{3}}{2 \sqrt{x}}$

Let $q \left(x\right) = {x}^{2} - 4$, then $q ' \left(x\right) = 2 x$

$f ' \left(x\right) = \frac{\left(\frac{\sqrt{3}}{2 \sqrt{x}}\right) \left({x}^{2} - 4\right) - \sqrt{3 x} \left(2 x\right)}{{x}^{2} - 4} ^ 2$

Since we only care about the derivative at $x = 3$, we shall leave simplifying the expression.

$f ' \left(3\right) = \frac{\left(\frac{\sqrt{3}}{2 \sqrt{3}}\right) \left({3}^{2} - 4\right) - \sqrt{3 \left(3\right)} \left(2 \left(3\right)\right)}{{3}^{2} - 4} ^ 2 = - \frac{31}{50}$