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

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If y^x = e^(y-x) ,then show that dy/dx=(logey)²/(logy)

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

then prove that

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

x^{y} = y^{x}

\frac{d y}{d x} = \frac{y (x log y - y)}{x (y log x - x)}

A = ⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & {log}_{x} y & {log}_{x} z {log}_{y} x & 4 & {log}_{y} z {log}_{z} x & {log}_{z} y & 6 \end{matrix} ⎤ ⎥ ⎥ ⎦

log x = {log}_{e} x

log y = {log}_{e} y

log z = {log}_{e} z

e^{x} + e^{y} = e^{x + y}

\frac{d y}{d x} = - e^{y - x}

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

\frac{d y}{d x} = \frac{- e^{y}}{1 + x^{2}}

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