# Understanding partial derivative involving 3 variables

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).

• There is sign error at $z^2=\dots$ Oct 3, 2018 at 22:09
• @Bernard edited! Oct 3, 2018 at 22:16

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 $$\;\dfrac{\partial z}{\partial y}$$.
• I am rather new to this so I don't really follow your train of thought. Does this mean that I can simply substitute the values $(2,2,1)$ and $(2,2,-1)$ into the equation as stated, such that for point $(2,2,1)$, the partial derivative is $\frac{4}{3}$ and for $(2,2,-1)$, it is $-\frac{4}{3}$? Oct 3, 2018 at 22:41
• Isn't $\frac{\partial z}{\partial x} = \frac{2}{3}$ at $(2, 2, 1)$ and $\frac{\partial z}{\partial x} = -\frac{2}{3}$ at $(2, 2, -1)$ based on your own calculations? Oct 4, 2018 at 9:45
• Do you mean For $\frac{\partial z}{\partial x} = \frac{x}{\sqrt{3}\sqrt{x^2 + y^2 - 5}}$ When $z = 1: \frac{\partial z}{\partial x} = \sqrt{\frac23}$ When $z = -1, \frac{\partial z}{\partial x} = -\sqrt{\frac23}$ Is this what you meant? Oct 3, 2018 at 22:21