# Math Help - partial derivative question

1. ## partial derivative question

I need some clarification on how to get the $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$ on:

$x^2 +y^2+z^2 = 3xyz$

I know that the solution is $\frac{\partial z}{\partial x} = \frac{3yz-2x}{2z-3xy}$ and $\frac{\partial z}{\partial y} = \frac{3xz-2y}{2z-3xy}$ but I have no idea how they got to it.

2. Let's do $\frac{\partial{z}}{\partial{x}}$

We can use implicit differentiation letting y be a constant.

$x^{2}+y^{2}+z^{2}=3xyz$

$2x+2z\frac{dz}{dx}=3xy\frac{dz}{dx}+3yz$

$2z\frac{dz}{dx}-3xy\frac{dz}{dx}=3yz-2x$

$\frac{dz}{dx}(2z-3xy)=3yz-2x$

$\frac{dz}{dx}=\frac{3yz-2x}{2z-3xy}$

3. Originally Posted by lllll
I need some clarification on how to get the $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$ on:

$x^2 +y^2+z^2 = 3xyz$

I know that the solution is $\frac{\partial z}{\partial x} = \frac{3yz-2x}{2z-3xy}$ and $\frac{\partial z}{\partial y} = \frac{3xz-2y}{2z-3xy}$ but I have no idea how they got to it.
Use implicit differentiation (z van be treated as an implicit function of x and y):

Differentiate both sides partially with resepct to x:

$2x + 2z \frac{\partial z}{\partial x} = 3yz + 3xy \frac{\partial z}{\partial x}$.

The chain rule has been used on the left hand side to differentiate z^2 and the product rule has been used on the right to differentiate the product.

Now re-arrange the result to make $\frac{\partial z}{\partial x}$ the subject.

Same idea when getting the partial derivative with respect to y.

Note that you can re-arrange the given rule to make z the subject and then take the partial derivative with respect to x in the usual way.