Question

What is the derivative of ${\sin }^{-1}\left(\frac{2x}{1+x^2}\right)$ with respect to ${\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)$.

• A. $1$
• B. $2$
• C. $-1$
• D. $4$

Answer 1

The correct option is A

$1$

Step 1: Solving ${\sin }^{-1}\left(\frac{2x}{1+x^2}\right)$

Let, $u={\sin }^{-1}\left(\frac{2x}{1+x^2}\right)$

Put , $x=\tan \varnothing$

$\Rightarrow \varnothing ={\tan }^{-1}x$

Now, solving $u$.

$$\begin{gathered} \Rightarrow u={\sin }^{-1}\left(\frac{2\tan \varnothing }{1+{\tan }^2\varnothing }\right) \\ ={\sin }^{-1}\left(\frac{2\tan \varnothing }{se{c}^2\varnothing }\right) \\ ={\sin }^{-1}\left(\frac{2{\displaystyle \frac{\sin \varnothing }{\cos \varnothing }}}{{\displaystyle \frac{1}{{\cos }^2\varnothing }}}\right) \\ ={\sin }^{-1}\left(2\sin \varnothing \cos \varnothing \right) \\ ={\sin }^{-1}\left(\sin 2\varnothing \right)\mathbf{(}\mathbf{\because }\mathbf{sin}\mathbf{}\mathbf{2}\mathbf{\varnothing }\mathbf{}\mathbf{=}\mathbf{}\mathbf{2}\mathbf{}\mathbf{sin}\mathbf{\varnothing }\mathbf{}\mathbf{cos}\mathbf{\varnothing }\mathbf{)} \\ =2\varnothing \\ =2{\tan }^{-1}x \end{gathered}$$

Step 2: Finding $\frac{du}{dx}$

Now finding, $\frac{du}{dx}=\frac{d\left(2{\tan }^{-1}x\right)}{dx}$

$\Rightarrow \frac{du}{dx}=\frac{2}{1+x^2}$

Step 3: Solving ${\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)$

Let, $v={\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)$

Substituting ,$x=\tan \varnothing$

$\Rightarrow \varnothing ={\tan }^{-1}x$

Now, solving $v$

$$\begin{gathered} \Rightarrow v={\cos }^{-1}\left(\frac{1-{\tan }^2\varnothing }{1+{\tan }^2\varnothing }\right) \\ ={\cos }^{-1}\left(\frac{1-{\tan }^2\varnothing }{se{c}^2\varnothing }\right)\mathbf{}\left(1+{\tan }^2\varnothing ={\sec }^2\varnothing \right) \\ ={\cos }^{-1}\left(\frac{1}{se{c}^2\varnothing }-\frac{{\tan }^2\varnothing }{se{c}^2\varnothing }\right) \\ ={\cos }^{-1}\left({\cos }^2\varnothing -{\sin }^2\varnothing \right) \\ ={\cos }^{-1}\left(\cos 2\varnothing \right) \\ =2\varnothing \\ =2{\tan }^{-1}x \end{gathered}$$

Step 4: Finding $\frac{dv}{dx}$

Now finding, $\frac{dv}{dx}=\frac{d\left(2{\tan }^{-1}x\right)}{dx}$

$\Rightarrow \frac{dv}{dx}=\frac{2}{1+x^2}$

Step 5: Finding the derivative

We need, $\frac{du}{dv}=\frac{\left({\displaystyle \frac{du}{dx}}\right)}{\left({\displaystyle \frac{dv}{dx}}\right)}$

$\begin{array}{rcl}\frac{du}{dv}& =& \frac{\left({\displaystyle \frac{2}{1+x^2}}\right)}{\left({\displaystyle \frac{2}{1+x^2}}\right)}\\ & =& 1\end{array}$

Hence, the derivative of ${\sin }^{-1}\left(\frac{2x}{1+x^2}\right)$ with respect to ${\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)$ is $1$.

Thus, option(A) is the correct answer.