Question

If $y={\tan }^{-1}\frac{\left(\sqrt{\left(1+x^2\right)}-1\right)}{x}$ then $y\text{'}\left(1\right)=$

• A. $\frac{1}{4}$
• B. $\frac{-1}{4}$
• C. $\frac{1}{2}$
• D. $\frac{-1}{2}$

Answer 1

The correct option is A

$\frac{1}{4}$

Explanation for the correct option:

Finding the value of $y\text{'}\left(1\right)$:

Given,

$y={\tan }^{-1}\frac{\left(\sqrt{\left(1+x^2\right)}-1\right)}{x}$

Now put ,

$$\begin{gathered} x=\tan \theta \\ \theta ={\tan }^{-1}x \end{gathered}$$

Then,

$\begin{array}{rcl}y& =& {\tan }^{-1}\frac{\left(\sqrt{\left(1+{\tan }^2\theta \right)}-1\right)}{\tan \theta }\\ & =& {\tan }^{-1}\frac{\left(\sqrt{se{c}^2\theta }-1\right)}{\tan \theta }\mathbf{[}\mathbf{\because }\mathbf{1}\mathbf{+}\mathbf{t}\mathbf{a}{\mathbf{n}}^{\mathbf{2}}\mathbf{\theta }\mathbf{}\mathbf{=}\mathbf{}\mathbf{s}\mathbf{e}{\mathbf{c}}^{\mathbf{2}}\mathbf{\theta }\mathbf{}\mathbf{]}\\ & =& {\tan }^{-1}\frac{sec\theta -1}{\tan \theta }\\ & =& {\tan }^{-1}\left(\frac{{\displaystyle \frac{1}{\cos \theta }}-1}{{\displaystyle \frac{\sin \theta }{\cos \theta }}}\right)\mathbf{}\mathbf{}\mathbf{[}\mathbf{\because }\mathbf{}\mathbf{s}\mathbf{e}\mathbf{c}\mathbf{}\mathbf{\theta }\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{1}}{\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{}\mathbf{\theta }}\mathbf{}\mathbf{,}\mathbf{}\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{}\mathbf{\theta }\mathbf{=}\frac{\mathbf{}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{\theta }}{\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{}\mathbf{\theta }}\mathbf{]}\\ & =& {\tan }^{-1}\left(\frac{1-\cos \theta }{\sin \theta }\right)\\ & =& {\tan }^{-1}\left(\frac{2{\sin }^2{\displaystyle \frac{\theta }{2}}}{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}\right)\mathbf{[}\mathbf{\because }\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{}\mathbf{\theta }\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}\mathbf{}\mathbf{-}\mathbf{}\mathbf{2}\mathbf{}\mathbf{s}\mathbf{i}{\mathbf{n}}^{\mathbf{2}}\frac{\mathbf{\theta }}{\mathbf{2}}\mathbf{}\mathbf{,}\mathbf{}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{}\mathbf{\theta }\mathbf{}\mathbf{=}\mathbf{2}\mathbf{}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{}\frac{\mathbf{\theta }}{\mathbf{2}}\mathbf{}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{}\frac{\mathbf{\theta }}{\mathbf{2}}\mathbf{}\mathbf{]}\\ & =& {\tan }^{-1}\left(\tan \frac{\theta }{2}\right)\\ & =& \frac{\theta }{2}\\ & =& \frac{{\tan }^{-1}x}{2}\end{array}$

Now differentiate with respect to $x$.

$$\begin{gathered} \frac{dy}{dx}=\frac{1}{2}\frac{1}{1+x^2}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{[}\mathbf{\because }\frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}\mathbf{t}\mathbf{a}{\mathbf{n}}^{\mathbf{-}\mathbf{1}}\mathbf{x}\mathbf{}\mathbf{=}\frac{\mathbf{1}}{\mathbf{1}\mathbf{+}{\mathbf{x}}^{\mathbf{2}}}\mathbf{}\mathbf{]} \\ y\text{'}\left(1\right)=\frac{1}{2}\frac{1}{1+{\left(1\right)}^2} \\ =\frac{1}{4} \end{gathered}$$

Hence, the correct option is A.