This question could possibly be posted in a different forum but I think the crux of the issue is an incorrect mathematical derivation.
This Wikipedia website provides equations for converting from ECEF to Geodetic coordinates. When I compute $\frac{\partial z}{\partial\phi}$ at the equator, I'm getting a very large value: 6335439.327292461 meters per degree latitude, but I expect something more along the lines of 111,000 meters per degree latitude.
I'm starting with $$ z(\phi) = \left(\frac{b^2}{a^2}N(\phi) + h\right)\sin(\phi) $$ where $b$ is the polar radius of the Earth, $a$ is the equatorial radius of the Earth, $N$ is the prime vertical radius of curvature, and $\phi$ is latitude.
The equation I'm using for $N$ is $$ N = \frac{a^2}{\sqrt{a^2\cos^2(\phi) + b^2\sin^2(\phi)}} $$
And here's how I've computed the partial: $$ \frac{\partial z(\phi)}{\partial \phi} = \frac{b^2}{a^2}\frac{\partial N(\phi)}{\partial \phi}\sin(\phi) + \left(\frac{b^2}{a^2}N(\phi) + h\right)\cos(\phi) $$ with $$ \frac{\partial N(\phi)}{\partial \phi} = \frac{N^3}{a^4}(a^2-b^2)\cos(\phi)\sin(\phi) $$
Evaluating this at $\phi=0$ yields: $$ \frac{\partial N(\phi)}{\partial \phi}\Bigg|_{\phi=0} = \frac{b^2}{a^2}N(0\phi) + h $$ which is roughly the value I have listed.
Did I compute $\frac{\partial N(\phi)}{\partial \phi}$ incorrectly? Or am I interpreting the partial incorrectly? I thought it should tell me the meter change per degree.
Are you getting meters per radian longitude, instead of degree? That figure of 6.3 million meters seems to be about right for that (because it's pretty close to the Earth's radius.) If you multiply your figure by $\pi/180$, you get 110,574.3 m, which seems close to what you were expecting.