# How do you find the first and second derivative of ln sqrt (x^2-4)?

Apr 25, 2017

$\frac{\mathrm{df}}{\mathrm{dx}} = \frac{x}{{x}^{2} - 4}$ and $\frac{{d}^{2} f}{{\mathrm{dx}}^{2}} = - \frac{{x}^{2} + 4}{{x}^{2} - 4} ^ 2$

#### Explanation:

We can use chain rule here, for which first differentiate w.r.t. $\sqrt{{x}^{2} - 4}$ and ten with respect to $\left({x}^{2} - 4\right)$ and finally w.r.t. $x$.

As such, as $f \left(x\right) = \ln \left(\sqrt{{x}^{2} - 4}\right)$

$\frac{\mathrm{df}}{\mathrm{dx}} = \frac{1}{\sqrt{{x}^{2} - 4}} \times \frac{1}{2 \sqrt{{x}^{2} - 4}} \times 2 x$

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

It is apparent that for second derivative, we can use quotient formula and hence

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

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

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