Question

If $y={\left({\tan }^{-1}x\right)}^2$, then ${\left(x^2+1\right)}^2{y}_2+2x\left(x^2+1\right)y_1$ is equal to

• A. $4$
• B. $0$
• C. $2$
• D. $1$

Answer 1

The correct option is C

$2$

Explanation for the correct option.

Step 1. Find the first derivative.

Differentiate both sides of the equation $y={\left({\tan }^{-1}x\right)}^2$ with respect to $x$.

$\begin{array}{rcl}{y}_1& =& \frac{d}{dx}\left({\left({\tan }^{-1}x\right)}^2\right)\\ & \Rightarrow & y_1=2{\tan }^{-1}x\frac{1}{1+x^2}\mathbf{[}\mathbf{\because }\mathbf{}\frac{\mathbf{d}}{\mathbf{dx}}\left({\tan }^{-1}x\right)\mathbf{=}\frac{\mathbf{1}}{\mathbf{1}\mathbf{+}{\mathbf{x}}^{\mathbf{2}}}\mathbf{]}\\ & \Rightarrow & \left(1+x^2\right)y_1=2{\tan }^{-1}x...\left(1\right)\end{array}$

Step 2. Find the value of the given expression.

Differentiate both sides of the equation $y_1=\frac{2{\tan }^{-1}x}{1+x^2}$ with respect to $x$.

$$\begin{gathered} y_2=\frac{d}{dx}\left(\frac{2{\tan }^{-1}x}{1+x^2}\right) \\ \Rightarrow y_2=\frac{2\times {\displaystyle \frac{1}{1+x^2}}\left(1+x^2\right)-2{\tan }^{-1}x\left(2x\right)}{{\left(1+x^2\right)}^2} \\ \Rightarrow {\left(1+x^2\right)}^2{y}_2=2-4x{\tan }^{-1}x \\ \Rightarrow {\left(1+x^2\right)}^2{y}_2+2x\left(2{\tan }^{-1}x\right)=2 \\ \Rightarrow {\left(1+x^2\right)}^2{y}_2+2x\left(1+x^2\right)y_1=2\left[\mathrm{Using}\mathrm{equation}1\right] \end{gathered}$$

Hence, the correct option is C.