Question

If $y=ta{n}^{-1}\frac{\sqrt{1+x^2}-1}{x}$, then

• A. $y^{\prime }\left(1\right)=1$
• B. $y^{\prime }\left(1\right)=\frac{1}{4}$
• C. $y^{\prime }\left(1\right)=0$
• D. $y^{\prime }\left(1\right)$ does not exist.

Answer 1

The correct option is B $y^{\prime }\left(1\right)=\frac{1}{4}$

Putting $x=tan\theta$, we have
$\frac{\sqrt{1+x^2}-1}{x}=\frac{sec\theta -1}{tan\theta }=\frac{1-cos\theta }{sin\theta }=tan\frac{\theta }{2}$
$y=\frac{\theta }{2}=\frac{1}{2}ta{n}^{-1}x$
$y^{\prime }\left(x\right)=\frac{1}{2(1+x^2)}\Rightarrow y^{\prime }\left(1\right)=\frac{1}{4}$