Question

If e^(x+y)=y^(x) then (dy)/(dx)=?

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Answer 1

$\huge\mathfrak{solution}$

$\star \: {e}^{x + y} = {y}^{x}$

Take log both sides :

$\implies \: log( {e})^{x + y} = log(y) {}^{ x } \\ \\ \implies \: (x + y) log(e) = x log(y) \\ \\ \implies \: x + y = x log( y)$

Now , differentiate w.r.t x both sides ,

$\implies \: \: \frac{d}{dx} (x + y) = \frac{d}{dx} (x \: log(y) ) \\ \\ \implies \: 1 + \frac{dy}{dx} = x \frac{d}{dx} log(y) + log(y) \frac{d}{dx} x \\ \\ \implies \: 1 + \frac{dy}{dx} = x( \frac{1}{y} ) \frac{dy}{dx} + log(y) (1) \\ \\ \implies \: \frac{dy}{dx} (1 - \frac{x}{y} ) = log(y) - 1 \\ \\ \implies \: \frac{dy}{dx} = \frac{y( log(y) - 1) }{y - x}$

Answer 2

Answer:

Correct Question:-

$If \: {e}^{x+y} = xy \: \: \: then \: \frac{dy}{dx} =?$

Solution:-

$log \: {e}^{(x+y)} = log \: x.y \\ (x+y) \: log \: e=log \: x.y \\ x+y=log \: x+log \: y \\ 1+ \frac{dy}{dx} = \frac{1}{x} + \frac{1}{y} \: \frac{dy}{dx} \\ \frac{dy}{dx} - \frac{1}{y} \: \frac{dy}{dx} = \frac{1}{x} - 1 \\ \frac{dy}{dx} (1- \frac{1}{y} )= (\frac{1}{x} -1) \\ \frac{dy}{dx} =( \frac{y - 1}{y} )=( \frac{1-x}{x} ) \\ \frac{dy}{dx} = (\frac{y}{y-1} ) (\frac{1-x}{x} ) \\ \frac{dy}{dx} = \frac{y}{x} ( \frac{1-x}{y-1} )$