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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!

1 Answers 1

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Note: This answer was posted before the correction. However, the same idea still works.

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.

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    @Akshit: The question "Find $Z_x$ and $Z_y$" implicitly assumes (no pun intended) that $z$ is implicitly a function of $x,y$.2016-09-27