Reciprocal of Partial Derivatives Involving Cartesian to Spherical Coordinate

multivariable-calculuspartial derivative

So for conversion of spherical to Cartesian coordinate,

$$ {x = \rho\sin \phi\cos \theta}, {y = \rho\sin \phi\sin \theta}, {z = \rho\cos \phi}
$$

$$\rho = \sqrt{x^2 + y^2 + z^2} \to \frac{\partial \rho}{\partial x} = \frac{x}{\sqrt{x^2 + y^2 + z^2}} = \sin\phi \cos\theta$$

$$ x = \rho\sin \phi cos \theta \to \frac{\partial x}{\partial \rho} = \sin\phi \cos\theta$$

But isn't the following true?

$$
\frac{\partial \rho}{\partial x} = \frac{1}{\frac{\partial x}{\partial \rho} }
$$

Then which of my following steps above is wrong?

This question is a continuation from my previous question here.

Best Answer

You're comparing $\left(\frac{\partial\rho}{\partial x}\right)_{y,\,z}$ (meaning the partial derivative is defined with $y,\,z$ held constant) to $\left(\frac{\partial x}{\partial\rho}\right)_{\theta,\,\phi}^{-1}$ (meaning $\theta,\,\phi$ are held constant instead). If you relate Cartesian and polar coordinates in two dimensions, you'll see a simpler example of the importance of like-for-like comparisons in this respect. See also the triple product rule and Maxwell relations.