Question:
If \(y=(tan^{-1}x)^2\),show that \((x^2+1)^2y_2+2x(x^2+1)y_1=2\)
If \(y=(tan^{-1}x)^2\),show that \((x^2+1)^2y_2+2x(x^2+1)y_1=2\)
Updated On: Sep 13, 2023
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Solution and Explanation
The given relationship is,\(y=(tan^{-1}x)^2\)
Then,\(y_1=2tan^{-1}x\frac{d}{dx}(tan^{-1}x)\)
\(⇒y_1=2tan^{-1}x.\frac{1}{1+x^2}\)
\(⇒(1+x^2)y_1=2tan^{-1}x\)
Again differentiating with respect to \(x\) on both sides,we obtain
\((1+x^2)y_2+2xy_1=2(\frac{1}{1+x^2})\)
\(⇒(1+x^2)^2y_2+2x(x^2+1)y_1=2\)
Hence,proved
Then,\(y_1=2tan^{-1}x\frac{d}{dx}(tan^{-1}x)\)
\(⇒y_1=2tan^{-1}x.\frac{1}{1+x^2}\)
\(⇒(1+x^2)y_1=2tan^{-1}x\)
Again differentiating with respect to \(x\) on both sides,we obtain
\((1+x^2)y_2+2xy_1=2(\frac{1}{1+x^2})\)
\(⇒(1+x^2)^2y_2+2x(x^2+1)y_1=2\)
Hence,proved
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