document.write( "Question 765940: what is the solution to:\r
\n" ); document.write( "\n" ); document.write( "implicit differentiation\r
\n" ); document.write( "\n" ); document.write( "e^x + e^y = e^x+y \n" ); document.write( "

Algebra.Com's Answer #466525 by MaartenRU(13) $\"\"$ $\"About$

You can put this solution on YOUR website!
I will assume you mean $\"e%5Ex%2Be%5Ey=e%5E%28x%2By%29\"$ , and not $\"e%5Ex%2Be%5Ey=e%5Ex%2By\"$ , because that would require $\"e%5Ey=y\"$ , which has no real solution.\r
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\n" ); document.write( "\n" ); document.write( "Alright, we now have
\n" ); document.write( " $\"e%5Ex%2Be%5Ey=e%5E%28x%2By%29\"$
\n" ); document.write( "We will first substitute $\"u=e%5Ey\"$ :
\n" ); document.write( " $\"e%5Ex%2Bu=u%2Ae%5Ex\"$
\n" ); document.write( "Now we want to separate u from the rest of the equation.
\n" ); document.write( " $\"e%5Ex%2Bu-u%2Ae%5Ex=0\"$
\n" ); document.write( "Factor out u:
\n" ); document.write( " $\"e%5Ex%2Bu%281-e%5Ex%29=0\"$
\n" ); document.write( " $\"u%28e%5Ex-1%29=e%5Ex\"$
\n" ); document.write( " $\"u=%28e%5Ex%29%2F%28e%5Ex-1%29\"$ \r
\n" ); document.write( "\n" ); document.write( "So putting the substitution back gives us
\n" ); document.write( " $\"e%5Ey=%28e%5Ex%29%2F%28e%5Ex-1%29\"$
\n" ); document.write( "And so:
\n" ); document.write( " $\"y=ln%28%28e%5Ex%29%2F%28e%5Ex-1%29%29\"$ \r
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\n" ); document.write( "\n" ); document.write( "Of course this requires that the denominator is not zero, which translates to x not equal to zero. For any other x, this is the y that satisfies the equation. \n" ); document.write( "

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