# How do you find the second derivative of -ln(x-(x^2+1)^(1/2))?

May 7, 2017

$\frac{- x}{{x}^{2} + 1} ^ \left(\frac{3}{2}\right)$

#### Explanation:

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

To find the first derivative:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 1}{x - {\left({x}^{2} + 1\right)}^{\frac{1}{2}}} \left[\frac{d}{\mathrm{dx}} \left(x - {\left({x}^{2} + 1\right)}^{\frac{1}{2}}\right)\right]$

Using the chain rule again:

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

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

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

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

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

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

Then:

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = - \frac{1}{2} {\left({x}^{2} + 1\right)}^{- \frac{3}{2}} \left(2 x\right)$

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = - {\left({x}^{2} + 1\right)}^{- \frac{3}{2}} \left(x\right)$

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \frac{- x}{{x}^{2} + 1} ^ \left(\frac{3}{2}\right)$