When I was studying multivariable calculus, there would be questions like
Find $\frac{\partial}{\partial r} (r^2 \sin\theta)$ and evaluate it at $r = 2$.
And basically an acceptable answer (in the multivariable course that I was doing), would be to just "symbolically" differentiate and get $\frac{\partial}{\partial r} (r^2 \sin\theta) = 2r\sin\theta$ and then $4\sin\theta$ when $r = 2$. And usually such functions wouldn't even have a domain specified but usually one would assume that such a function was stated in polar coordinates.
But in learning multivariable analysis and basic differential geometry where I'm forced to use calculus more rigorously, I learnt that there is only one real definition for partial derivatives in $\mathbb{R}^n$.
Definition: Let $U \subseteq \mathbb{R}^n$ be an open set and let $f : U \to \mathbb{R}$. The $i^{\text{th}}$ partial derivative of $f$ at $p \in U$ is defined to be $$\frac{\partial f}{\partial x^i} (p) = \lim_{t \to 0} \frac{f(p + te_i)-f(p)}{t}$$ where $e_i$ is the $i$-th basis vector for $\mathbb{R}^n$ in other words $e_i = (0, \dots, 1, \dots, 0)$ with $1$ in the $i$-th position.
Now suppose I have a scenario where I want to calculate the partial derivative of a function defined in polar coordinates.
To be more precise let $P = (0, \infty) \times \mathbb{R}$ (where we want to think of $P$ as being polar coordinates), let $V = (0, \infty) \times (-\pi, \pi)$ and let $f : V \to \mathbb{R}$ be defined by $f(r, \theta) = r^2\sin\theta$. Now the only way I think it would be possible to calculate $\frac{\partial f}{\partial r} (2, 0)$ would be to find a diffeomorphism $F$ from $V$ to an open subset $U$ of $\mathbb{R}^2$ and then evaluate the partial derivative of the composite $f \circ F^{-1}$.
It turns out that the function $F : (0, \infty) \times \mathbb{R} \to \mathbb{R}^2$ defined by $F(r, \theta) = (r\cos\theta, r\sin\theta)$ gives a diffeomorphism from $V$ to $U = \{(x, y) \in \mathbb{R}^2 \ | \ x > 0, y \in \mathbb{R} \} \subseteq \mathbb{R}^2$. I then think that $$\frac{\partial f}{\partial r} (2, 0) = \frac{\partial (f \circ F^{-1})}{\partial x} (F(2, 0))$$
Am I correct in saying that? If so why was it so easy to take and evaluate partial derivatives in multivariable calculus courses when the rigorous approach seems a lot more difficult?