# Thread: [SOLVED] Partial derivation

1. ## [SOLVED] Partial derivation

I'm new at this, so...
$\displaystyle yln{x}\ln{y}dx+dy=0$

then, making it a function:

$\displaystyle f(x,y)=\frac{dy}{dx}= -yln{x}\ln{y}$

$\displaystyle \frac{\partial{f}}{\partial{y}}= ?$

any help (and further explanations) will be great...

thanks in advance ^^

2. Hello,
Originally Posted by kirbyiwaki
I'm new at this, so...
$\displaystyle yln{x}\ln{y}dx+dy=0$

then, making it a function:

$\displaystyle f(x,y)=\frac{dy}{dx}= -yln{x}\ln{y}$

$\displaystyle \frac{\partial{f}}{\partial{y}}= ?$

any help (and further explanations) will be great...

thanks in advance ^^
In other words, you're asked for the partial derivative of $\displaystyle -y \ln(x) \ln(y)$, with respect to y.

This means that you have to consider x as a constant. Thus $\displaystyle \ln(x)$ will be treatead as a constant. If you can't see it, substitute it by k.

$\displaystyle g(y)=-k \cdot y\ln(y)$

Can you find $\displaystyle g'(y)$ ?

3. OMG, that was easy xD... looks like the $\displaystyle \ln$ scared me >_>... thanks n__n"