How do you differentiate # 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^4-2x^2+2)}#

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

#\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)}#