Stuck on this one also
Given that y=y(x,z) is defined implicitly by the relation
$\displaystyle y^3 + z\sqrt{x} - ln(xz^{2}-y) = \frac{16}{x^2} +24 $
compute $\displaystyle \frac{dy}{dz} $ at the point (4,-3,1)
Thanks!
the code for a partial is \partial y
$\displaystyle 3y^2\frac{\partial y}{\partial z}+\sqrt{x}-\frac{2xz-\frac{\partial y}{\partial z}}{xz^2-y}=0$
eval @ (4,-3,1)
$\displaystyle 3(-3)^2\frac{\partial y}{\partial z}+\sqrt{4}-\frac{2(4)(1)-\frac{\partial y}{\partial z}}{(4)(1)^2-(-3)}=0$
$\displaystyle 27 \frac{\partial y}{\partial z} +2 -\frac{8-\frac{\partial y}{\partial z}}{7}=0$
$\displaystyle 189 \frac{\partial y}{\partial z} +14 -8 +\frac{\partial y}{\partial z}=0 \iff 190 \frac{\partial y}{\partial z} =-6 \iff \frac{\partial y}{\partial z}=\frac{-3}{95} $