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Question
If log y = tan−1 x, show that (1 + x2)y2 + (2x − 1) y1 = 0 ?
Solution
Here,
\[\log y = \tan^{- 1} x\]
\[\text { Differentiating w . r . t . x, we get }\]
\[\frac{1}{y} \times y_1 = \frac{1}{1 + x^2}\]
\[ \Rightarrow \left( 1 + x^2 \right) y_1 = y\]
\[ \Rightarrow \left( 1 + x^2 \right) y_2 + 2x y_1 = y_1 \]
\[ \Rightarrow \left( 1 + x^2 \right) y_2 + 2x y_1 - y_1 = 0\]
\[ \Rightarrow \left( 1 + x^2 \right) y_2 + \left( 2x - 1 \right) y_1 = 0\]
Hence proved.
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f(x) = 3x2 + 6x + 8, x ∈ R
