Question

For the function $y={\tan }^{-1}x,(1+x^2)y_2=$ ____

• A. $-2x{y}_1$
• B. $x{y}_1$
• C. $2x{y}_1$
• D. $-x{y}_1$

Answer 1

The correct option is A $-2x{y}_1$

$y={\tan }^{-}1\left(x\right)$

therefore

$x=\tan \left(y\right)\Rightarrow \frac{dx}{dx}={\sec }^2\left(y\right)y_1\Rightarrow y_1={\cos }^2\left(y\right)$

differentiating on both sides

$y_1=-2\cos \left(y\right)\sin \left(y\right)y_2{y}_2=-\frac{1}{2\cot \left(y\right)}$

therefore

$(1+x^2)y_2=-2x{y}_1$