How do you differentiate $g (x) = \sqrt{arctan (x^{2} - 1)}$ $g(x) = sqrtarctan(x^2-1)$ ?

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Answer 1 · 2016-05-28T07:36:51

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

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

$\frac{d f (u)}{d x} = \frac{d f}{d u} \cdot \frac{d u}{d x}$ $\frac{df(u)}{dx}=\frac{df}{du}\cdot \frac{du}{dx}$

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

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

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

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

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

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

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

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

How do you differentiate g(x)=√arctan(x2−1) g(x) = sqrtarctan(x^2-1) ?