Question

$find \: \frac{d^{2}y}{dx^{2} } \\ \\ \\ y = \tan(x + y)$

Chapter: Second Order Derivative
Standard: 12

Answer:
$- \frac{2(1 + y^{2}) }{y^{5} }$

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Answer 1

Hi...☺

Here is your answer...✌

y = tan(x + y) -----(1)

Differentiating both sides w.r.t. x
We get

$\frac{dy}{dx} = { \sec}^{2} (x + y) \times (1 + \frac{dy}{dx} ) \\ \\ \frac{dy}{dx} =[1 + { \tan }^{2} (x + y) ] \times (1 + \frac{dy}{dx} ) \\ \\ \frac{dy}{dx} = (1 + {y}^{2} ) \times (1 + \frac{dy}{dx} ) \: \: \: \: \:[from \: (1)] \\ \\ \frac{dy}{dx} = 1 + \frac{dy}{dx} + {y}^{2} + {y}^{2} \frac{dy}{dx} \\ \\ - {y}^{2} \frac{dy}{dx} = 1 + {y}^{2} \\ \\ \frac{dy}{dx} = \frac{ - (1 + {y}^{2}) }{ {y}^{2} } \: \: \: \: \: .....(2)$
Again differentiating both sides w.r.t. x
We get

$\frac{ {d}^{2}y }{d {x}^{2} } = - \frac{2y\frac{dy}{dx} \times {y}^{2} - (1 + {y}^{2} ) 2y \frac{dy}{dx} }{{ ({y}^{2} ) }^{2} } \\ \\ \frac{ {{d} }^{2} y}{d {x}^{2} } =- \frac{dy}{dx} ( \frac{2 {y}^{3} - 2y - 2 {y}^{3} }{ {y}^{4} } ) \\ \\ \frac{ {d}^{2}y }{d {x}^{2} }=- \frac{dy}{dx}( \frac{ - 2y}{ {y}^{4} } ) \\ \\ \frac{{d }^{2}y }{d {x}^{2} } = \frac{dy}{dx} \times \frac{2}{ {y}^{3} } \\ \\ \frac{ {d}^{2} y}{d {x}^{2} } = \frac{ - (1 + {y}^{2}) }{ {y}^{2} } \times \frac{2}{ {y}^{3}} \: \: \: \: \: [ \: from \: (2) \: ] \\ \\ \frac{ {d}^{2}y }{d {x}^{2} } = \frac{ - 2(1 + {y}^{2}) }{ {y}^{5} }$