1. Partial Derivatives

Find $\displaystyle \frac{ \partial z }{ \partial y }$ where $\displaystyle z$ is defined implicitly as a function of $\displaystyle x$ and $\displaystyle y$ by:

$\displaystyle sin(xyz) = x + 2y + 3z$

2. $\displaystyle \frac{\partial}{\partial y}\ [sin(xyz) = x+2y+3z]$

$\displaystyle cos(xyz)xz \frac{\partial z}{\partial y}\ = 0+2+3\frac{\partial z}{\partial y}\$

$\displaystyle \frac{\partial z}{\partial y}\ (xzcos(xyz)-3) = 2$

$\displaystyle \frac{\partial z}{\partial y}\ = \frac{2}{xzcos(xyz)-3}$

Could someone please check my work?

3. Originally Posted by pberardi
$\displaystyle \frac{\partial}{\partial y}\ [sin(xyz) = x+2y+3z]$

$\displaystyle cos(xyz)xz \frac{\partial z}{\partial y}\ = 0+2+3\frac{\partial z}{\partial y}\$ *

[snip]
If z is an implicit function of x and y then the partial derivative of xyz with respect to y is $\displaystyle xz + xy \frac{\partial z}{\partial y}$.

Therefore the left hand side of the line marked * should be $\displaystyle \cos (xyz) \left(xz + xy \frac{\partial z}{\partial y} \right)$.