# Find the partial derivative of a sphere with equation $x^2+y^2+z^2=4$

We have a sphere with the following equation:

$x^2+y^2+z^2=4$

We seek to find the partial derivative, with respect to $x$, of this equation. We think of this equation as a function of three variables

$f(x,y,z)=4$

From this equation, we can conclude the following:

$\frac{df}{dx}=0$

Using the chain rule, we have:

$0=f_x\frac{dx}{dx}+f_y\frac{dy}{dx}+f_z\frac{dz}{dx}$

$0=(2x)(1)+(2y)(0)+(2z)\frac{dz}{dx}$

Since we are treating $y$ as a constant, from the equation above, we can solve the very last variable.

$0=2x+(2z)\frac{dz}{dx}$

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

I am quite confused regarding the treatment of $z$ and $y$. I've always assumed that when seeking a partial derivative with respect to a certain variable, we treat the other variables as constants. Can it be said that as in the example above, we were treating $z$ as a constant all along? If yes, should $\frac{dz}{dx}$ evaluate to $0$?

• Citation : << Find the partial derivative of a sphere with equation ...>>. I known what is the partial derivative of a FUNCTION of several variables. I am eager to know what is the partial derivative of a SPHERE and also of an EQUATION. – JJacquelin Feb 9 '16 at 8:54
• Alright, perhaps I used bad terminology, but do you have a helpful input? – TheValars Feb 9 '16 at 15:43
• I don't say that the terminology is bad. But I cannot understand it . Is it literally the initial wording of the problem ? – JJacquelin Feb 9 '16 at 16:07

## 2 Answers

The way I understand it is you have the equation $$x^2+y^2+z^2=4$$ which is equivalent to $$f(x,y)=z=\pm \sqrt{4-x^2-y^2},$$ therefore $$\frac{\partial{f}}{\partial{x}}=\pm \frac{x}{\sqrt{4-x^2-y^2}}$$ Perhaps more context on where this question comes from could help clarify things.

We can treat other variabes as constants only when we calculate $\frac{\partial f}{\partial x}$ or $\frac{\partial f}{\partial y}$ or $\frac{\partial f}{\partial z}$ as to the frst order.
When we calculate $\frac{df}{dx}$ or $\frac{df}{dy}$ or $\frac{df}{dz}$ themselves, we should not treat $x$ or $y$ or $z$ as constants.
In general $\frac{df}{dx} \ne 0 ,\, \frac{df}{dy} \ne 0 ,\, \frac{df}{dx} \ne 0$ and $\frac{dz}{dx}\ne 0\,(f=z)$ .

There are two standpoints.
Standpoint 1
If we regard $x,\,y,\,z$ as independent variables, then $$\frac{\partial z}{\partial x} = \frac{\partial z}{\partial y} =0$$ Under the constraint $x^2+y^2+z^2=4$, by differentiating the both sides wrt x we get
$$\frac{dx^2}{dx} + \frac{\partial y^2}{\partial y}\frac{dy}{dx} + \frac{\partial z^2}{\partial z}\frac{dz}{dx}=0$$ $$x + y\frac{dy}{dx} + z\frac{dz}{dx}=0$$ \begin{alignat}{2} \frac{dz}{dx} &=&& -\frac{x}{z} - \frac{y}{z}\frac{dy}{dx} \\ &=&& \mp \frac{x}{\sqrt{4-x^2-y^2}} \mp \frac{y}{\sqrt{4-x^2-y^2}} \frac{dy}{dx} \end{alignat}

Standpoint 2
If we regard $z$ as a function of $x,\,y$ $$z=\pm \sqrt{4-x^2-y^2}$$ as Kuifje wrote, then $$\frac{\partial z}{\partial x} \ne 0 ,\quad \frac{\partial z}{\partial y} \ne 0$$ \begin{alignat}{2} \frac{dz}{dx} &=&& \frac{\partial z}{\partial x} + \frac{\partial z}{\partial y} \frac{dy}{dx} \\ &=&& \mp \frac{x}{\sqrt{4-x^2-y^2}} \mp \frac{y}{\sqrt{4-x^2-y^2}} \frac{dy}{dx} \end{alignat}

Two standpoints give the same expression for $\frac{dz}{dx}$.