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Q. Find `dy/dx` , where `x^y` = ```e^(x-y)` .

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Q. Find `dy/dx` , where `x^y` = ```e^(x-y)` .

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Before taking the derivative with respect to x, isolate the y. To do so, take the natural logarithm of both sides.


On each side of the equation, apply the power rule of logarithm which is `ln a^m=m lna` .


Note that ln e=1 .


Bring together the terms with y on one side of the equation. So, add both sides by y.



Factor out the GCF at the left side.


And, divide both sides by 1 + lnx.



Now that the variable y is isolated, let's take the derivative with respect to x.


To take the derivative, apply quotient rule `(u/v)'=(v*u'-u*v')/v^2` .





Hence, `(dy)/(dx)=lnx/(1+lnx)^2` .

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