Question:
If $y=\left(\tan ^{-1} x\right)^{2}$, show that $\left(x^{2}+1\right)^{2} y_{2}+2 x\left(x^{2}+1\right) y_{1}=2$
Solution:
The given relationship is $y=\left(\tan ^{-1} x\right)^{2}$
Then,
$y_{1}=2 \tan ^{-1} x \frac{d}{d x}\left(\tan ^{-1} x\right)$
$\Rightarrow y_{1}=2 \tan ^{-1} x \cdot \frac{1}{1+x^{2}}$
$\Rightarrow\left(1+x^{2}\right) y_{1}=2 \tan ^{-1} x$
Again differentiating with respect to $x$ on both the sides, we obtain
$\left(1+x^{2}\right) y_{2}+2 x y_{1}=2\left(\frac{1}{1+x^{2}}\right)$
$\Rightarrow\left(1+x^{2}\right)^{2} y_{2}+2 x\left(1+x^{2}\right) y_{1}=2$
Hence, proved.
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.