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In a similar spirit, whenever one has an $n$-dimensional smooth manifold $M$, and (locally) one has $n$ smooth coordinate functions $x^1,\dots,x^n: M \to {\bf R}$ on this manifold, whose differentials $dx^1,\dots,dx^n$ form a basis of the cotangent space at every point $p$ of the manifold $M$, then (locally at least) there is a unique "dual basis" of derivations $\partial_1,\dots,\partial_n$ on $C^\infty(M)$ with the property $\partial_i x^j = \delta_i^j$ for $i,j=1,\dots,n$. (By the way, proving this claim is an excellent exercise for someone who really wants the modern foundations of differential geometry.)

In a similar spirit, whenever one has an $n$-dimensional smooth manifold $M$, and (locally) one has $n$ smooth coordinate functions $x^1,\dots,x^n: M \to {\bf R}$ on this manifold, whose differentials $dx^1,\dots,dx^n$ form a basis of the cotangent space at every point $p$ of the manifold $M$, then (locally at least) there is a unique "dual basis" of derivations $\partial_1,\dots,\partial_n$ on $C^\infty(M)$ with the property $\partial_i x^j = \delta_i^j$ for $i,j=1,\dots,n$. (By the way, proving this claim is an excellent exercise for someone who really wants to understand the modern foundations of differential geometry.)

In a similar spirit, whenever one has an $n$-dimensional smooth manifold $M$, and (locally) one has $n$ smooth coordinate functions $x^1,\dots,x^n: M \to {\bf R}$ on this manifold, whose differentials $dx^1,\dots,dx^n$ form a basis of the cotangent space at every point $p$ of the manifold $M$, then (locally at least) there is a unique "dual basis" of derivations $\partial_1,\dots,\partial_n$ on $C^\infty(M)$ with the property $\partial_i x^j = \delta_i^j$ for $i,j=1,\dots,n$. (By the way, proving this claim is an excellent exercise for someone who really wants to understand such basic concepts in modern differential geometry such as derivations and manifolds.)

In a similar spirit, whenever one has an $n$-dimensional smooth manifold $M$, and (locally) one has $n$ smooth coordinate functions $x^1,\dots,x^n: M \to {\bf R}$ on this manifold, whose differentials $dx^1,\dots,dx^n$ form a basis of the cotangent space at every point $p$ of the manifold $M$, then (locally at least) there is a unique "dual basis" of derivations $\partial_1,\dots,\partial_n$ on $C^\infty(M)$ with the property $\partial_i x^j = \delta_i^j$ for $i,j=1,\dots,n$. (By the way, proving this claim is an excellent exercise for someone who really wants the modern foundations of differential geometry.)

In a similar spirit, whenever one has an $n$-dimensional smooth manifold $M$, and (locally) one has $n$ smooth coordinate functions $x^1,\dots,x^n: M \to {\bf R}$ on this manifold, whose differentials $dx^1,\dots,dx^n$ form a basis of the cotangent space at every point $p$ of the manifold $M$, then (locally at least) there is a unique "dual basis" of derivations $\partial_1,\dots,\partial_n$ on $C^\infty(M)$ with the property $\partial_i x^j = \delta_i^j$ for $i,j=1,\dots,n$.

Now, traditionally, the derivation $\partial_i$ is instead denoted $\frac{\partial}{\partial x^i}$. But the notation is a bit misleading as it suggests that $\frac{\partial}{\partial x^i}$ only depends on the $i^{th}$ coordinate function $x^i$, when in fact it depends on the entire basis $x_1,\dots,x_n$ of coordinate functions. One can fix this by using more complicated notation, e.g., $\frac{\partial}{\partial x^i}|_{x^1,\dots,x^{i-1},x^{i+1},\dots,x^n}$, which informally means "differentiate with respect to $x^i$ while holding the other coordinates $x^1,\dots,x^{i-1},\dots,x^{i+1},\dots,x^n$ fixed". One sees this sort of notation for instance in thermodynamics. Of course, things are much simpler in the one-dimensional setting $n=1$; here, any coordinate function $x$ (with differential $dx$ nowhere vanishing) gives rise to a unique derivation $\frac{d}{dx}$ such that $\frac{d}{dx} x = 1$.

If one also had another coordinate $t$ on the same domain $M$, then one would have another differential $\frac{d}{dt}$ on $M$, which is related to the original differential $\frac{d}{dx}$ by the usual chain rule $$ \frac{d}{dt} f = \left(\frac{d}{dt} x\right) \left(\frac{d}{dx} f\right).$$ Again, for conceptual clarity, $t, x, f: M \to {\bf R}$ should all be viewed here as functions of a position variable $p \in M$, rather than being viewed as functions of each other.

In a similar spirit, whenever one has an $n$-dimensional smooth manifold $M$, and (locally) one has $n$ smooth coordinate functions $x^1,\dots,x^n: M \to {\bf R}$ on this manifold, whose differentials $dx^1,\dots,dx^n$ form a basis of the cotangent space at every point $p$ of the manifold $M$, then (locally at least) there is a unique "dual basis" of derivations $\partial_1,\dots,\partial_n$ on $C^\infty(M)$ with the property $\partial_i x^j = \delta_i^j$ for $i,j=1,\dots,n$. (By the way, proving this claim is an excellent exercise for someone who really wants to understand such basic concepts in modern differential geometry such as derivations and manifolds.)

Now, traditionally, the derivation $\partial_i$ is instead denoted $\frac{\partial}{\partial x^i}$. But the notation is a bit misleading as it suggests that $\frac{\partial}{\partial x^i}$ only depends on the $i^{th}$ coordinate function $x^i$, when in fact it depends on the entire basis $x^1,\dots,x^n$ of coordinate functions. One can fix this by using more complicated notation, e.g., $\frac{\partial}{\partial x^i}|_{x^1,\dots,x^{i-1},x^{i+1},\dots,x^n}$, which informally means "differentiate with respect to $x^i$ while holding the other coordinates $x^1,\dots,x^{i-1},\dots,x^{i+1},\dots,x^n$ fixed". One sees this sort of notation for instance in thermodynamics. Of course, things are much simpler in the one-dimensional setting $n=1$; here, any coordinate function $x$ (with differential $dx$ nowhere vanishing) gives rise to a unique derivation $\frac{d}{dx}$ such that $\frac{d}{dx} x = 1$.

If one also had another coordinate $t: M \to {\bf R}$ on the same domain $M$, then one would have another differential $\frac{d}{dt}$ on $M$, which is related to the original differential $\frac{d}{dx}$ by the usual chain rule $$ \frac{d}{dt} f = \left(\frac{d}{dt} x\right) \left(\frac{d}{dx} f\right).$$ Again, for conceptual clarity, $t, x, f: M \to {\bf R}$ should all be viewed here as functions of a position variable $p \in M$, rather than being viewed as functions of each other.