In general, you can manipulate total derivatives like fractions, but you can’t do the same with partial derivatives.
Cross-posted from here.
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If you have a function $f(x,y)$ where $x=x(t)$ and $y=y(t)$ are themselves functions of a parameter $t,$ and you blindly cancel out differentials, then you can get to incorrect statements like
$$\begin{align*} \require{cancel}\dfrac{\partial f}{\partial t} = \dfrac{\partial f}{\cancel{\partial x}} \cdot \dfrac{\cancel{\partial x}}{\partial t} = \dfrac{\partial f}{\cancel{\partial y}} \cdot \dfrac{\cancel{\partial y}}{\partial t}, \quad {\color{red}\times} \end{align*}$$
whereas what’s actually true is
$$\begin{align*} \dfrac{\partial f}{\partial t} = \dfrac{\partial f}{\partial x} \cdot \dfrac{\partial x}{\partial t} + \dfrac{\partial f}{\partial y} \cdot \dfrac{\partial y}{\partial t}. \quad {\color{green}\checkmark} \end{align*}$$
You can’t cancel because the $\partial f$’s in the numerators of $\dfrac{\partial f}{\partial t},$ $\dfrac{\partial f}{\partial x},$ $\dfrac{\partial f}{\partial y}$ all mean different things.
- The $\partial f$ in the numerator of $\dfrac{\partial f}{\partial t},$ represents the change in $f$ attributed to the change in $t.$
- The $\partial f$ in the numerator of $\dfrac{\partial f}{\partial x}$ represents the change in $f$ attributed to the change in $x.$
- The $\partial f$ in the numerator of $\dfrac{\partial f}{\partial y}$ represents the change in $f$ attributed to the change in $y.$
But in single-variable calculus, you’re working exclusively with functions that have only one input variable. And if you have a function $f(x)$ where $x=x(t)$ is itself a function of a parameter $t,$ then it’s true that
$$\begin{align*} \dfrac{\partial f}{\partial t} = \dfrac{\partial f}{\partial x} \cdot \dfrac{\partial x}{\partial t}. \end{align*}$$
The above is conventionally written with “total” derivative symbols ($\mathrm d$ means “total”, $\partial$ means “partial”) since the change attributed to the single variable is the same as the total change of the function.
$$\begin{align*} \dfrac{\mathrm df}{\mathrm dt} = \dfrac{\mathrm df}{\mathrm dx} \cdot \dfrac{\mathrm dx}{\mathrm dt} \end{align*}$$
So in general, you can manipulate total derivatives ($\mathrm d$) like fractions, but you can’t do the same with partial derivatives ($\partial$).
$$\begin{align*} \require{cancel} \textrm{valid:} \quad &\dfrac{\mathrm df}{\mathrm dt} = \dfrac{\mathrm df}{\cancel{\mathrm dx}} \cdot \dfrac{\cancel{\mathrm dx}}{\mathrm dt} \quad {\color{green}\checkmark} \\[5pt] \textrm{NOT valid:} \quad &\dfrac{\partial f}{\partial t} = \dfrac{\partial f}{\cancel{\partial x}} \cdot \dfrac{\cancel{\partial x}}{\partial t} \quad {\color{red}\times} \end{align*}$$
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