document.write( "Question 1208324: if y ^(y) = e ^(x - y) find (dy)/(dx) at x = 1 \n" ); document.write( "
Algebra.Com's Answer #846673 by math_tutor2020(3817)\"\" \"About 
You can put this solution on YOUR website!

\n" ); document.write( "y^y = e^(x-y)
\n" ); document.write( "Ln( y^y ) = Ln( e^(x-y) )
\n" ); document.write( "y*Ln(y) = (x-y)*Ln(e)
\n" ); document.write( "y*Ln(y) = (x-y)*1
\n" ); document.write( "y*Ln(y) = x-y\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Apply the implicit derivative to both sides
\n" ); document.write( "y*Ln(y) = x-y
\n" ); document.write( "y'*Ln(y) + y*(1/y)*y' = 1-y'
\n" ); document.write( "y'*Ln(y) + y' + y' = 1
\n" ); document.write( "y'*Ln(y) + 2*y' = 1
\n" ); document.write( "y'*( Ln(y) + 2 ) = 1
\n" ); document.write( "y' = 1/( Ln(y) + 2 )\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "We'll come back to this later.
\n" ); document.write( "Return to the original equation.
\n" ); document.write( "Plug in x = 1 to find its paired y value.
\n" ); document.write( "y^y = e^(x-y)
\n" ); document.write( "y^y = e^(1-y)
\n" ); document.write( "At first glance this equation looks impossible to solve by hand.
\n" ); document.write( "But through trial-and-error you would find that y = 1 is a solution.
\n" ); document.write( "There may be other solutions.
\n" ); document.write( "y^y = e^(1-y)
\n" ); document.write( "1^1 = e^(1-1)
\n" ); document.write( "1 = e^(0)
\n" ); document.write( "1 = 1\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "So,
\n" ); document.write( "y' = 1/( Ln(y) + 2 )
\n" ); document.write( "y' = 1/( Ln(1) + 2 )
\n" ); document.write( "y' = 1/( 0 + 2 )
\n" ); document.write( "y' = 1/2
\n" ); document.write( "
\n" ); document.write( " \n" ); document.write( "
\n" );