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

[randy17]

Homework Statement

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

Homework Equations

xy^2+xlnx=4y

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?

 

[jeppetrost]

The function ln, as you might know, isn't defined for values less than zero. Hence it is meaningless to speak of a function value, let alone a derivative there.

By the way, for this to make sense to me, you have to have an equal sign somewhere. Like xy^2+xlnx = 0 or something - else it's just a value, not anything to define a function y(x).

 

[randy17]

By the way, for this to make sense to me, you have to have an equal sign somewhere. Like xy^2+xlnx = 0 or something - else it's just a value, not anything to define a function y(x).

Sorry, the eq is actually xy^2+xlnx=4y. Thank you for pointing out the mistake, I'll edit it out.

 

[TheoMcCloskey]

I get a slightly different answer. Check your derivative of the first term on the left.