Question

Class _12 Differentiation of implicit functions​

Answer 1

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$\bf\Huge\red{\mid{\overline{\underline{ ANSWER }}}\mid }$

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$\Large\fbox{\color{purple}{QUESTION}}$

$log( \sqrt{ {x}^{2} + {y}^{2} } ) = { \tan}^{ (- 1)} \frac{y}{x}$

We have to find ...

$\implies \frac{dy}{dx}$

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$\Large\fbox{\color{purple}{ SOLUTION }}$

$\implies log( \sqrt{ {x}^{2} + {y}^{2} } ) = { \tan}^{ (- 1)} \frac{y}{x} \\ \implies \frac{1}{ \sqrt{ {x}^{2} + {y}^{2} } } \times \frac{1}{2} \times \frac{2x + 2y (\frac{dy}{dx}) }{ \sqrt{ {x}^{2} + {y}^{2} } } = \frac{1}{1 + ( { \frac{y}{x} })^{2} } \times \frac{x \frac{dy}{dx} - y \times 1 }{ {x}^{2} } \\ \implies \frac{1}{ ({x}^{2} + {y}^{2}) } \times x + y \frac{dy}{dx} = \frac{ {x}^{2} }{( {x}^{2} + {y}^{2} )} \times \frac{x \times \frac{dy}{dx} - y}{ {x}^{2} } \\ \implies x + y \frac{dy}{dx} = x\frac{dy}{dx} - y \\ \implies y \frac{dy}{dx} - x \frac{dy}{dx} = - y - x \\ \implies \frac{dy}{dx} = - \frac{y + x}{y - x}$

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