Is a reciprocal partial derivative equal to the reciprocal of the partial derivative?

Question

In particular, I have $$\frac{\partial r}{\partial x} = \frac{\partial (x^2 + y^2 + z^2)^{1/2}}{\partial x} = \frac{x}{(x^2 + y^2 + z^2)^{1/2}}= \frac{x}{r}$$ does this mean $$\frac{\partial x}{\partial r} = \frac{r}{x}$$ ?
If it is, can we safely assume that this is so for all partial derivatives?

Answer 1

Of course, an equality is an equality (so this statement is general). The equality $ \frac{\partial x(r,y,z)}{\partial r} = \frac{r}{x}$ is also true.

You can easy proof that in this case using $ x = \sqrt{r^2 - y^2 - z^2}$. You will get: $$ \frac{\partial x}{\partial r} = \frac{2r}{2{\sqrt{r^2 - y^2 - z^2}}} = \frac{r}{x} $$.

More general, for a intuitive proof, let as consider $f = f(x,y,z)$ (it can easy be generalized to a function with $n$ variables. I will also try to simplify the notation).

$$ \frac{\partial f}{\partial x} = \lim_{\Delta x \to 0 } \frac{ f(x + \Delta x , y, z) - f(x,y,z)}{\Delta x} =\lim_{\Delta x \to 0 } \frac{\Delta f}{\Delta x} $$

$$ \frac{\partial x}{\partial f} = \lim_{\Delta f \to 0 } \frac{ x(f + \Delta f , y, z) - x(f,y,z)}{\Delta f} =\lim_{\Delta f \to 0 } \frac{\Delta x}{\Delta f} $$

From the second equality we get: $$\frac{1}{\frac{\partial x}{\partial f}} = \lim_{\Delta f \to 0 } \frac{\Delta f}{\Delta x} =\lim_{\Delta x \to 0 } \frac{\Delta f}{\Delta x} = \frac{\partial f}{\partial x} $$

(In the last step we used that for $\Delta f \to 0 $ and $ y,z $ fixed also $\Delta x \to 0 $)