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Question
If y = (cot−1 x)2, prove that y2(x2 + 1)2 + 2x (x2 + 1) y1 = 2 ?
Solution
Here,
\[y = \left( \cot^{- 1} x \right)^2 \]
\[\text { Differentiating w . r . t . x, we get }\]
\[ y_1 = 2co t^{- 1} x \times \frac{- 1}{1 + x^2} = \frac{- 2co t^{- 1} x}{1 + x^2}\]
\[\text { Differentiating again w . r . t . x, we get }\]
\[ y_2 = \frac{2 + 4x \cot^{- 1} x}{\left( 1 + x^2 \right)^2}\]
\[ \Rightarrow y_2 = \frac{2}{\left( 1 + x^2 \right)^2} + \frac{2x \times 2 \cot^{- 1} x}{\left( 1 + x^2 \right)\left( 1 + x^2 \right)}\]
\[ \Rightarrow y_2 = \frac{2}{\left( 1 + x^2 \right)^2} - \frac{2x y_1}{\left( 1 + x^2 \right)}\]
\[ \Rightarrow \left( 1 + x^2 \right)^2 y_2 = 2 - 2x y_1 \left( 1 + x^2 \right)\]
\[ \Rightarrow \left( 1 + x^2 \right)^2 y_2 + 2x y_1 \left( 1 + x^2 \right) = 2\]
Hence proved.
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