Question

If $y=se{c}^{-1}\left(\frac{\left(x+1\right)}{\left(x-1\right)}\right)+{\sin }^{-1}\left(\frac{\left(x-1\right)}{\left(x+1\right)}\right)$, then $\frac{dy}{dx}=$

• A. $1$
• B. $0$
• C. $\frac{x-1}{x+1}$
• D. $\frac{x+1}{x-1}$

Answer 1

The correct option is B

$0$

Explanation for the correct answer:

Find the value of $\frac{dy}{dx}$:

Given,

$$\begin{gathered} y=se{c}^{-1}\left(\frac{\left(x+1\right)}{\left(x-1\right)}\right)+{\sin }^{-1}\left(\frac{\left(x-1\right)}{\left(x+1\right)}\right) \\ \Rightarrow y={\cos }^{-1}\left(\frac{\left(x-1\right)}{\left(x+1\right)}\right)+{\sin }^{-1}\left(\frac{\left(x-1\right)}{\left(x+1\right)}\right)\mathbf{}\mathbf{}\mathbf{}\mathbf{[}\mathbf{\because }\mathbf{s}\mathbf{e}{\mathbf{c}}^{\mathbf{-}\mathbf{1}}\mathbf{\left(}\mathbf{a}\mathbf{\right)}\mathbf{}\mathbf{=}\mathbf{}\mathbf{c}\mathbf{o}{\mathbf{s}}^{\mathbf{-}\mathbf{1}}\mathbf{\left(}\frac{\mathbf{1}}{\mathbf{a}}\mathbf{\right)}\mathbf{]} \\ \Rightarrow y=\frac{\mathrm{\pi }}{2}\mathbf{[}\mathbf{\because }\mathbf{}\mathbf{c}\mathbf{o}{\mathbf{s}}^{\mathbf{-}\mathbf{1}}\mathbf{x}\mathbf{}\mathbf{+}\mathbf{}\mathbf{s}\mathbf{i}{\mathbf{n}}^{\mathbf{-}\mathbf{1}}\mathbf{x}\mathbf{}\mathbf{=}\frac{\mathbf{\pi }}{\mathbf{2}}\mathbf{]} \end{gathered}$$

Differentiate $y$ with respect to $x$, we get

$\frac{dy}{dx}=0\left[\because \frac{d\left(\mathrm{constant}\right)}{\mathrm{dx}}=0\right]$

Hence, the correct option is B.