Question 1208324: if y ^(y) = e ^(x - y) find (dy)/(dx) at x = 1
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
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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