Understanding partial derivative involving 3 variables

Question

I am new to partial derivative and I need some help in understanding if what I have done so far is correct.

Let $S$ be the surface given by $x^2 + y^2 - 3z^2 = 5$

I want to calculate the partial derivative:

$\frac{\partial z}{\partial x}$ at the point $(2,2,1)$ and $(2,2,-1)$

This is what I have done:

$x^2 + y^2 - 3z^2 = 5$

$z^2 = \frac{x^2 + y^2 - 5}{3}$

$z = \pm \sqrt\frac{x^2 + y^2 - 5}{3}$

$\frac{\partial z}{\partial x} = \frac{\frac{1}{2}(x^2 + y^2 - 5)^{-\frac12}(2x)}{\sqrt3}$

$\frac{\partial z}{\partial x} = \frac{2x}{2\sqrt{3}\sqrt{x^2 + y^2 - 5}}$

$\frac{\partial z}{\partial x} = \frac{x}{\sqrt{3}\sqrt{x^2 + y^2 - 5}}$

But I am unsure of how to continue after this, and how to use the points (2,2,1) and (2,2,-1).

Answer 1

Use differential calculus with the equation of the surface: differentiating both sides yields $$2x\,\mathrm dx+2y\,\mathrm dy-6z\,\mathrm dz=0,$$ whence $$\mathrm dz=\frac{x\,\mathrm dx+y\,\mathrm dy}{3z}.$$ Now $\;\dfrac{\partial z}{\partial x}$ is the coefficient of $\mathrm dx$, and similarly for $\;\dfrac{\partial z}{\partial y}$.

Answer 2

I'm also not expert partial derivative but I have thought like that: your final result is a global result. So you can find desired partial derivative just put x and y values into the found equations. Z values are only affect the sign of the result. For a positive z values, your solution is fully correct but for a negative z value, you should add a minus sign to the equation when square rooting z value in your solution.