$\displaystyle f(x,y)=ln(4x-3y)$, (5,-7)

So, to do this I just find $\displaystyle \frac{\partial f}{\partial x}$ and $\displaystyle \frac{\partial f}{\partial y}$ right?

If so, then:

$\displaystyle \frac{\partial f}{\partial x}=\frac{4}{4x-3y}$

$\displaystyle \frac{\partial f}{\partial y}=\frac{-3}{4x-3y}$

And then:

$\displaystyle f(5,-7)=\frac{4}{4(5)-3(-7)}i-\frac{3}{4(5)-3(-7)}j=>\frac{4}{41}i-\frac{3}{41}j$

Is this correct? Thanks.