What does "find the dx/dy of (eq inside) for x>0" mean?

[randy17]

1. The problem statement, all variables and given/known data

find dy/dx for the function xy^2+xlnx=4y for x>o

2. Relevant equations

xy^2+xlnx=4y

3. The attempt at a solution

I found the dy/dx to be "(y^2+lnx+1)/(4-x)=dx/dy"
What is it asking me to do with the x>0? Do they want me to plug in a value greater than 0 for x and solve for y? or what?

 

[pka]

1. The problem statement, all variables and given/known data

find dy/dx for the function xy^2+xlnx=4y for x>o

Your first step is:
\(\displaystyle \left( {{y^2} + 2xyy'} \right) + \left( {\ln (x) + 1} \right) = 4y'\)
Now solve for \(\displaystyle y'\) and you and done.

The only reason to state \(\displaystyle x>0\) is that it is the domain of \(\displaystyle \ln(x)\).

 

[HallsofIvy]

You need to be a lot more careful. In the title of you post you have "dx/dy" while in the statement of the problem you have "dy/dx". Later in your solution you say 'I found the dy/dx to be "(y^2+lnx+1)/(4-x)=dx/dy'". Do you see the difference? If I were your teacher I would point that out the first few times, then start deducting points. You need to pay more attention to what you are doing.

As pka says, the only reason for the restriction "x> 0" is that ln(x) is defined only for positive x.