# Partial derivative of an implicit function

I am trying to solve this question (sorry if the translation is a bit vague):

$Z(x,y)$ is an implicit function of $x$ and $y$ given in the form of

$$x^3z^2+\frac{2}{9}y^2\sin(z) = xyz$$

in the neighborhood of $x=2, y=\pi, z=\frac{\pi}{6}$.

Find $Z_x$ and $Z_y$ at $(x,y) = (2,\pi)$

I know how to partially/totally differentiate, and I know how to find the derivative of a single-variable implicit function. How do I combine it to solve this? I have the solution in front of me but I can't understand it. Thanks!

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I assume the given equation is $$x^3z^2 + \frac{2}{9}y^2\sin z = \frac{\pi^2}{3}.$$ Let's now regard $z = Z(x,y)$ as a function of $x$ and $y$ and implicitly (partially) differentiate with respect to $x$: $$\frac{\partial}{\partial x}\left( x^3z^2 \right) + \frac{\partial}{\partial x}\left(\frac{2}{9}y^2\sin z \right) = \frac{\partial}{\partial x}\left(\frac{\pi^2}{3} \right).$$ That is, $$3x^2 z^2 + x^3 \frac{\partial}{\partial x}(z^2) + \frac{2}{9}y^2\frac{\partial}{\partial x}(\sin z) = 0,$$ so $$3x^2 z^2 + x^3 (2z)\, Z_x + \frac{2}{9}y^2(\cos z)\,Z_x = 0.$$ Rearranging gives $$Z_x(x,y) = \frac{-3x^2z^2}{x^3(2z) + \frac{2}{9}y^2\cos z}.$$ Now, we know that when $x = 2$ and $y = \pi$, we have $z = \frac{\pi}{6}$. Plugging these values in, we find that \begin{align} Z_x(2, \pi) & = \frac{-3(2^2)(\frac{\pi}{6})^2}{2^3(\frac{2\pi}{6}) + \frac{2}{9}\pi^2\cos\frac{\pi}{6}} \\ & = \frac{ \frac{-\pi^2}{3} }{ \frac{8\pi}{3} + \frac{\pi^2\sqrt{3}}{9} }. \end{align} Simplifying this expression gives something like $$Z_x(2,\pi) = \frac{-3\pi}{24 + \pi\sqrt{3}},$$ though I might have made a calculation error along the way.
The $Z_y$ case is similar and is left to you.
Can you explain going from $x^3 \frac{\partial}{\partial x}(z^2)$ to $x^3 (2z)\, Z_x$? Also, I thought that $Z_x$ is the partial derivative of $z$ with respect to $x$ – yotamoo Jan 23 '12 at 6:45
Sure. This is just the chain rule: $\frac{\partial}{\partial x}(z(x,y)^2) = 2z(x,y)\,\frac{\partial z}{\partial x}$, and by definition, $Z_x = \frac{\partial z}{\partial x}$. – Jesse Madnick Jan 23 '12 at 12:39
Multivariable calculus is a place where variables can sometimes be a little confusing. For this reason, it's often easier to think in terms of functions rather than variables. But at any rate, the notation $z = Z(x,y)$ is much like the notation $y = f(x)$, and I imagine you'd be comfortable with writing either $\frac{dy}{dx}$ or $f'(x)$ to mean the same thing. The same goes for $Z_x = \frac{\partial z}{\partial x}$. – Jesse Madnick Jan 23 '12 at 12:42