Question

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Answer 1

Question :

If y= $e{}^{\frac{y}{x}}$

Prove that :

$\frac{dy}{dx} = \frac{y {}^{2} }{x(y - x)}$

Solution :

$y = e{}^{\frac{y}{x}}$

Now take log on both sides

$\implies log(y) = \dfrac{y}{x} log_{e}(e)$

$\implies log(y) = \dfrac{y}{x}$

Now differinate with respect to x

$\implies \frac{1}{y} \times \frac{dy}{dx} = \frac{1}{x} \times \frac{dy}{dx} + \frac{ - 1}{x {}^{2} } \times y$

$\implies \frac{dy}{dx} ( \frac{1}{y} - \frac{1}{x}) = \frac{ - y}{x {}^{2} }$

$\implies \frac{dy}{dx} ( \frac{x - y}{xy} ) = \frac{ - y}{x {}^{2} }$

$\implies \frac{dy}{dx} = \frac{ - y(yx)}{ x{}^{2}(x - y) }$

$\implies \dfrac{dy}{dx} = \dfrac{y {}^{2} }{x(y - x)}$

Hence proved!

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Some Formulas related to Differention:

$1) \frac{d( log(x)) }{dx} = \frac{1}{x}$

$2) \frac{d(x {}^{n}) }{dx} = nx {}^{n - 1}$

Answer 2

Answer:

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