We have a sphere with the following equation:


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


From this equation, we can conclude the following:


Using the chain rule, we have:



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



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 $0$?

  • 1
    $\begingroup$ 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. $\endgroup$ – JJacquelin Feb 9 '16 at 8:54
  • 1
    $\begingroup$ Alright, perhaps I used bad terminology, but do you have a helpful input? $\endgroup$ – TheValars Feb 9 '16 at 15:43
  • $\begingroup$ I don't say that the terminology is bad. But I cannot understand it . Is it literally the initial wording of the problem ? $\endgroup$ – JJacquelin Feb 9 '16 at 16:07

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 $\frac{dz}{dx}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.