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Log Function

If x^y=e^(y-x...

Question

If x^y=e^(y-x), prove that dy/dx=2-log x/(1-logx) ^2

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x^{y} = e^{x - y}, x \neq y

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

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

, then prove that

\frac{d y}{d x} = \frac{log x}{{[log e x]}^{2}}

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

then prove that

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

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

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