Would someone work through the derivative for this: (xy)^x = e Thanks.

Would someone work through the derivative for this: (xy)^x = e Thanks.

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logarithmic differentaition you want dy/dx I assume?

yeah, please

\[(xy)^x=e\]\[x\ln(xy)=1\]\[x(\ln x+\ln y)=1\]taking the derivative gives\[\ln x+\ln y+x(\frac1x+\frac{y'}y)=0\]\[\ln(xy)+1+\frac xyy'=0\]\[y'=-\frac yx(\ln(xy)+1)\]now note that\[y=\frac{e^{1/x}}x\]so we get\[y'=-\frac{e^{1/x}}{x^2}(\ln(e^{1/x})+1)\]\[y'=-\frac{e^{1/x}}{x^2}(\frac1x+1)\]

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