1. ## find y'

xlny=lnx^2

2. Originally Posted by johntuan
xlny=lnx^2
we differentiate implicitly, we use the product rule for the term on the left. i don't recall if i explained implicit differentiation to you or someone else recently, but that does not matter. implicit differentiation was explained several times on this site, you can search for "implicit differentiation" by using the search feature to find these posts.

$x \ln y = \ln x^2$

$\Rightarrow x \ln y = 2 \ln x$ ........i just rewrote the log in a more convenient form on the right so i don't have to use the chain rule on it

$\Rightarrow \ln y + \frac {x}{y}~y' = \frac {2}{x}$

Now just solve for $y'$

if you have any questions, say so

3. yes i understand so far now how do I solve for y'?

would it become y' = 2/x - lny all over x/y ?

4. Originally Posted by johntuan
yes i understand so far now how do I solve for y'?

would it become y' = 2/x - lny all over x/y ?
yes, and you can simplify that a bit as well. maybe $\frac {2y - xy \ln y}{x^2}$ would be more to your liking