I am a math hobbyist. I'm learning partial derivatives. All the examples I've come across deal with orthogonal coordinates. I am wondering what, if any, adjustments need to be made when taking partial derivatives in skew coordinates. When I've looked this question up on the web, the explanations were beyond me (e.g spoke of covariant and contravariant bases).
Let's say you have a simple function: $f(x,y) = 2x + 3y$. In an orthonormal coordinate system, the partial derivatives would be easy: $\frac{\partial f}{\partial x} = 3$, and $\frac{\partial f}{\partial y} = 2$. This depends on ones ability to consider $2x$ to be a constant when taking the partial with respect to $x$, and similarly to consider $3y$ to be constant when taking the partial with respect to $y$. But this wouldn't be the case in a skew coordinate system.
Say you have a two-dimensional coordinate system where the angle between the two axes is an acute angle $\phi$. (For example, the $y$-axis is rotated clockwise by $\frac{\pi}{2} - \phi$.) Then, when you move a distance, $h$, in the $x$ direction, you also move cos($\phi$) $\times $ $h$ in the $y$ direction. I've tried to work out what happens to partial derivatives as a result of this non-orthogonality between axes.
The formulation for the partial derivative with respect to $y$ in an orthonormal coordinate system is:
$$\lim_{h\to 0}\frac{f(x+h,y) - f(x,y)}{h}.$$
For the skew system I described above, since moving $h$ in the $x$ direction moves h cos($\phi$) $\times $ $h$ in the $y$ direction would this have to become:
$$\lim_{h\to 0}\frac{f(x+h,y+\cos(\phi) \times h) - f(x,y)}{h} ?$$
For $f(x,y) = 2x + 3y$, would $\dfrac{\partial f}{\partial y} = 2 + 3\cos(\phi)?$