Question

Which derivative and condition are correct for \[\tan^{-1}x\]?

पर्याय

• \[\frac{d}{dx}(\tan^{-1}x)=\frac{1}{\sqrt{1-x^2}},\quad |x|<1\]

• \[\frac{d}{dx}(\tan^{-1}x)=-\frac{1}{1+x^2},\quad x\in\mathbb{R}\]

• \[\frac{d}{dx}(\tan^{-1}x)=\frac{1}{1+x^2},\quad x\in\mathbb{R}\]

• \[\frac{d}{dx}(\tan^{-1}x)=\frac{1}{|x|\sqrt{x^2-1}},\quad |x|>1\]

Answer 1

For \[\tan^{-1}x\], the denominator is \[1+x^2\]. The condition is \[x\in\mathbb{R}\], and the derivative has no negative sign.