Question

If $u={\tan }^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)$ and $v={\tan }^{-1}x$, then $\frac{du}{dv}$is equal to

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

Answer 1

The correct option is C

$\frac{1}{4}$

Explanation for the correct option:

Step 1: Simplifying the given equation:

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

Put$x=\tan \theta$

$\begin{array}{rcl}u& =& {\tan }^{-1}\frac{(\surd (1+{\tan }^2\theta )–1)}{\tan \theta }\\ & =& {\tan }^{-1}\left(\frac{(sec\theta –1)}{\tan \theta }\right)\left[\because 1+ta{n}^{2}A=secA\right]\\ & =& {\tan }^{-1}\left(\frac{(1–\cos \theta )}{\sin \theta }\right)\\ & =& {\tan }^{-1}\tan \left(\frac{\theta }{2}\right)\left[\because \frac{1-cosA}{sinA}=tan\frac{A}{2}\right]\mathbf{}\mathbf{}\\ & =& \left(\frac{\theta }{2}\right)\\ & =& \frac{1}{2}{\tan }^{-1}x\left[\because x=tan\theta \right]\end{array}$

Step 2: Finding the derivative:

Differentiate $u$ and $v$ with respect to $x$

$\Rightarrow$$\frac{du}{dx}=\frac{1}{2(1+x^2)}...\left(1\right)\left[\because \frac{d}{dx}\left(ta{n}^{-1}x\right)=\frac{1}{1+x^2}\right]$

$$\begin{gathered} v=2{\tan }^{-1}x \\ \Rightarrow \frac{dv}{dx}=\frac{2}{1+x^2}...\left(2\right) \end{gathered}$$

Divide $\left(1\right)$ by $\left(2\right)$.

$$\begin{gathered} \frac{{\displaystyle \frac{du}{dx}}}{{\displaystyle \frac{dv}{dx}}}=\frac{{\displaystyle \frac{1}{2\left(1+x^2\right)}}}{{\displaystyle \frac{2}{1+x^2}}} \\ \Rightarrow \frac{du}{dx}=\frac{1}{4} \end{gathered}$$

Hence, the correct option is C.