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I want to derive the chain rule for partial derivatives, that is if $\gamma (t) = (x(t),y(t))$, then \begin{align*} \frac{\partial f\circ \gamma}{\partial t} (t) = \frac{\partial f}{\partial x} (x(t),y(t)) \frac{\partial x}{\partial t} (t) + \frac{\partial f}{\partial y} (x(t),y(t)) \frac{\partial y}{\partial t} (t). \end{align*} Here is what I have so far: \begin{align*} \frac{\partial f}{\partial t} (t)(h) & = Df (\gamma (t))(h) \\ & = Df (\gamma (t)) (D\gamma (t)(h)) \\ & = Df (\gamma (t))( D\gamma (t)(h_1\mathbf{e}_{1} ) + D\gamma (t)(h_2\mathbf{e}_{2} )) \\ & = Df (\gamma (t))\left(h_1 \begin{pmatrix} x' (t) \\ 0 \end{pmatrix} + h_2 \begin{pmatrix} 0 \\ y' (t) \end{pmatrix}\right) \\ & = h_1 D_1f (\gamma (t)) x' + h_2 D_2 f(\gamma (t))y'. \end{align*} Now I need to look at what $D_if(\gamma (t))$ looks like. We have \begin{align*} \left(\frac{\partial f}{\partial x_1} (\gamma (t))\right) = D_1 f(\gamma (t)) = \lim_{r \to 0} \frac{f(\gamma (t) + r\mathbf{e}_{1} ) - f(\gamma (t))}{r} & = \lim_{r \to 0} \frac{f(x(t) + r,y(t)) - f(x(t),y(t))}{r} \end{align*} Now I need show that this is $\frac{\partial f}{\partial x}(\gamma (t)) $ but I don't know how to formulate it as $x$ here is a variable and we have only defined partial derivatives in terms of coordinates of $\mathbb{R}^n$. For example, $x_1$ means with respect to the first coordinate as written above. Maybe what I have above is the definition of $\frac{\partial f}{\partial x} $?. Maybe we can count this $r$, per the above, as variation on $x(t)$ so it os okay? I am not sure. Could someone clarify this for me? Specifically, what is meant by $\frac{\partial f}{\partial x} $ where $x$ is variable rather than a direction?

Maths Wizzard
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    $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y},$ in the RHS of the identity you want to prove, are usual but uncomfortable notations for "partial derivatives of $f$ w.r.t. its first and second variables". The letters $x,y$ here are (usual but) completely arbitrary. You could as well denote them by $\partial_1f$ and $\partial_2f.$ And they are equal to what you wrote, i.e. to the directional derivatives of $f$ w.r.t. the first and second vectors of the canonical basis. Your proof is fine. – Anne Bauval May 24 '23 at 14:59
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    Just one remark: in the identity you want to prove, the LHS should not be written $\frac{\partial f}{\partial t} (t) $ but $\frac{d f\circ\gamma}{d t} (t) ,$ or $(f\circ\gamma)'(t).$ – Anne Bauval May 24 '23 at 15:02
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    Thank you all for the comments. I think I understand now. What confused me was the notation. – Maths Wizzard May 24 '23 at 15:11

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