# How do you differentiate  g(x) = sqrtarctan(x^2-1) ?

May 28, 2016

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

#### Explanation:

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

Applying chain rule,

$\setminus \frac{\mathrm{df} \left(u \setminus\right)}{\mathrm{dx}} = \setminus \frac{\mathrm{df}}{\mathrm{du}} \setminus \cdot \setminus \frac{\mathrm{du}}{\mathrm{dx}}$

Let $\arctan \left({x}^{2} - 1\right) = u$

$= \setminus \frac{d}{\mathrm{du}} \left(\setminus \sqrt{u}\right) \setminus \frac{d}{\mathrm{dx}} \left(\setminus \arctan \left({x}^{2} - 1\right)\right)$

We know,
$\setminus \frac{d}{\mathrm{du}} \left(\setminus \sqrt{u}\right) = \setminus \frac{1}{2 \setminus \sqrt{u}}$

and,
$\setminus \frac{d}{\mathrm{dx}} \left(\setminus \arctan \left({x}^{2} - 1\right)\right) = \setminus \frac{2 x}{{x}^{4} - 2 {x}^{2} + 2}$

So, $= \setminus \frac{1}{2 \setminus \sqrt{u}} \setminus \frac{2 x}{{x}^{4} - 2 {x}^{2} + 2}$

Substituting $\setminus : u = \setminus \arctan \left({x}^{2} - 1\right)$,we get

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

Simplifying it,
$= \setminus \frac{x}{\setminus \sqrt{\setminus \arctan \left({x}^{2} - 1\right)} \left({x}^{4} - 2 {x}^{2} + 2\right)}$