Im stuck, i forgot what do to next

[sbfp8]

The origional problem:

Find dy/dx

ln2xy=e^(x+y)


i made it too

[(x^-1) -ye^(x+y)] / (y^-1) - xe^(x+y) =dy/dx

 

[Gene]

Just to make sure what the problem is, use some more ()s.
ln(2xy)=e^(x+y) or
ln(2x)*y=e^(x+y) or
ln(2)*x*y=e^(x+y)
---------------
Gene

 

[pka]

If the problem is indeed \(\displaystyle \ln (2xy) = e^{x + y}\) then the next step is \(\displaystyle \frac{1}{x} + \frac{{y'}}{y} = \left( {1 + y'} \right)e^{x + y}\)

Now you must solve for y’.

 

[sbfp8]

im pretty sure thats how its supposed to be. it is written exactly the sameon my homework page as i wrote it. and the problem has to be done in terms of dy and dx. no primes. thank you tho.

 

[pka]

Well come now!
It is standard practice to write \(\displaystyle \L
y' = \frac{{dy}}{{dx}}\).

If you solve for y’ you find \(\displaystyle \L
\frac{{dy}}{{dx}}\) !

 

[sbfp8]

[(x^-1) -ye^(x+y)] / (y^-1) - xe^(x+y) =dy/dx

i made it to here, i just cant remember how to get rid of the fractions ( the x^-1 and y^-1) thats where i need help.