# Math Help - Partial Derivatives

1. ## Partial Derivatives

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

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

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

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

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

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

Could someone please check my work?

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

$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 $xz + xy \frac{\partial z}{\partial y}$.

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