# How do you find the derivative of  y= arctan(x/a)+ lnsqrt((x-a)/(x+a))?

Apr 10, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 a {x}^{2}}{{x}^{4} - {a}^{4}}$

#### Explanation:

Using the linearity of the derivative:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \arctan \left(\frac{x}{a}\right) + \frac{d}{\mathrm{dx}} \ln \sqrt{\frac{x - a}{x + a}}$

Now:

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

and using the properties of logarithms:

$\frac{d}{\mathrm{dx}} \ln \sqrt{\frac{x - a}{x + a}} = \frac{d}{\mathrm{dx}} \left(\frac{1}{2} \left(\ln \left(x - a\right) - \ln \left(x + a\right)\right)\right)$

$\frac{d}{\mathrm{dx}} \ln \sqrt{\frac{x - a}{x + a}} = \frac{1}{2} \left(\frac{1}{x - a} - \frac{1}{x + a}\right)$

$\frac{d}{\mathrm{dx}} \ln \sqrt{\frac{x - a}{x + a}} = \frac{1}{2} \left(\frac{\left(x + a\right) - \left(x - a\right)}{\left(x - a\right) \left(x + a\right)}\right)$

$\frac{d}{\mathrm{dx}} \ln \sqrt{\frac{x - a}{x + a}} = \frac{a}{{x}^{2} - {a}^{2}}$

Summing the two terms:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{a}{{x}^{2} + {a}^{2}} + \frac{a}{{x}^{2} - {a}^{2}}$

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