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

1 Answer
May 28, 2016

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

Explanation:

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

Applying chain rule,

\frac{df(u\)}{dx}=\frac{df}{du}\cdot \frac{du}{dx}

Let arctan (x^2-1)=u

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

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

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

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

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

=\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^4-2x^2+2)}