Question 1208324
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y^y = e^(x-y)
Ln( y^y ) = Ln( e^(x-y) )
y*Ln(y) = (x-y)*Ln(e)
y*Ln(y) = (x-y)*1
y*Ln(y) = x-y


Apply the implicit derivative to both sides
y*Ln(y) = x-y
y'*Ln(y) + y*(1/y)*y' = 1-y'
y'*Ln(y) + y' + y' = 1
y'*Ln(y) + 2*y' = 1
y'*( Ln(y) + 2 ) = 1
y' = 1/( Ln(y) + 2 )


We'll come back to this later.
Return to the original equation.
Plug in x = 1 to find its paired y value.
y^y = e^(x-y)
y^y = e^(1-y)
At first glance this equation looks impossible to solve by hand.
But through trial-and-error you would find that y = 1 is a solution. 
There may be other solutions.
y^y = e^(1-y)
1^1 = e^(1-1)
1 = e^(0)
1 = 1


So,
y' = 1/( Ln(y) + 2 )
y' = 1/( Ln(1) + 2 )
y' = 1/( 0 + 2 )
y' = 1/2
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