Question

if y= tan-1 x prove that ( 1 + x square ) y2 + 2xy 1 =0

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Answer 1

Step-by-step explanation:

We have,

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

Differentiating both sides w.r.t. x,

$\sf{ \implies \dfrac{dy}{dx} = \dfrac{1}{1 + {x}^{2} } }$

$\sf{ \implies y_{1} = \dfrac{1}{1 + {x}^{2} } }$

$\sf{ \implies (1 + {x}^{2} )y_{1} = 1 }$

Again, differentiating both sides w.r.t x,

$\sf{ \implies (1 + {x}^{2} )y_{2} + 2xy_{1} = 0 }$