Partial Derivatives

**janvdl** · December 28th 2008, 01:13 AM

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

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

Thanks in advance

**pberardi** · December 28th 2008, 02:47 AM

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

$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?

**mr fantastic** · December 28th 2008, 03:09 AM

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 $\cos (xyz) \left(xz + xy \frac{\partial z}{\partial y} \right)$ .

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