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Integrating Factor

ex-y=xy, then...

Question

$e^{x - y} = x^{y}, t h e n \frac{d y}{d x} =$

A

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B

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C

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D

$\frac{log~x}{(1+logx)^2}$

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Solution

The correct option is D
$\frac{log~x}{(1+logx)^2}$

$e^{x - y} = x^{y} \Rightarrow x - y = y l o g x ∴ y = \frac{x}{1 + l o g x}$

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Similar questions

x^{y} = e^{x - y}

\frac{d y}{d x} = \frac{l o g x}{(1 - l o g x)^{2}}

.

x^{y} = e^{x - y}

then prove that:

\frac{d y}{d x} = \frac{log x}{{(1 + log x)}^{2}}

x^{y} = e^{x - y}

, then show that

\frac{d y}{d x} = \frac{log x}{(1 + log x)^{2}}

x^{y} = e^{(x - y)}

\frac{d y}{d x} = \frac{log x}{(1 + log x)^{2}}

x^{y} = e^{x - y}

\frac{d y}{d x} = \frac{log x}{(1 + log x)^{2}}

.

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