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

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### 1 Answer

`x^y=e^(x-y)`

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

`lnx^y=lne^(x-y)`

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

Note that *ln e=1* .

`ylnx=x-y`

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

`y+ylnx=x-y+y`

`y+ylnx=x`

Factor out the GCF at the left side.

`y(1+lnx)=x`

And, divide both sides by 1 + lnx.

`(y(1+lnx))/(1+lnx)=x/(1+lnx)`

`y=x/(1+lnx)`

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

`d/(dx)y=d/(dx)(x/(1+lnx))`

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

`(dy)/(dx)=((1+lnx)(1)-x(0+1/x))/(1+lnx)^2`

`(dy)/(dx)=(1+lnx-x*1/x)/(1+lnx)^2`

`(dy)/(dx)=(1+lnx-1)/(1+lnx)^2`

`(dy)/(dx)=lnx/(1+lnx2)^2`

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