How does the following calculus notation read in plain English?

Question

What does the following notation read in plain English?

$df = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy$

As far as I can tell:

A vanishingly small change in $f(x,y)$ is equal to the sum of:

• change of $f(x,y)$ in the direction of $X$-axis multiplied by a vanishingly small change of the value of $x$

• change of $f(x,y)$ in the direction of $Y$-axis multiplied by a vanishingly small change of the value of $y$

Am I correct? If I am correct, What does this essentially mean?

Say it is a curve in a 3D space: Why the multiplication?

Answer 1

Your interpretation is not correct.

$$df=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy$$

is the derivative of $f$ in the direction of $(dx,dy)$ where $(dx,dy)$ is $any$ vector in what's called the "tangent space". The tangent space is a plane tangent to the function at a point. The origin of this plane is the point at which we took the derivative. Its vectors represent the changes that can me made to the "input" of the function. The output $df$ represents the change that would occur to the output of the function if the changes to the input were $(dx,dy)$.

It is perfectly valid to say $(dx,dy)=(2,-3)$ but that probably won't give us any useful information about the dynamics of $f$.

This is not exclusive to multivariable functions. Consider $f(x)=x^2$. If we take the derivative at $x=1$ we get $f'(1)=2(1)=2$.

It is a mistake to say that $2$ in this case is a number. The "2" in this case is a linear function that takes $dx$ to $2dx$. In other words, think of "2" as a one by one matrix. Then you will see that the total derivative is really the same derivative you have always known.

Answer 2

"Essentially" it means just what you say in your question (modulo the discussion in @JohnDouma 's answer).

Sometimes that is all you need to know to reason correctly in calculus and its applications.

What it means formally depends on the context. The differentials - those "vanishingly small changes" - have different definitions in different branches of mathematics.

(In your edit to the question you say $f(x,y)$ is a curve. That makes no sense to me - as $x$ and $y$ vary the graph in space is a surface.)

Answer 3

You're correct.

The reason it works like this in two dimensions is that, unlike in one, the "vanishingly small changes" you talk about can be along different paths, just like how you can approach a limit in multiple directions. So because of that, the change in a function with two inputs depends not just on the partial derivatives ($\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$) of the inputs, but also how much those inputs change. For example, the $df$ in your equation above will result in a different value based on the relative changes of $x$ and $y$ $-$ you can imagine the "vanishingly small changes" as a sort of vector, and as the vector rotates, even when its magnitude doesn't change, the individual components change, which in turn will affect $df$.

Answer 4

Intuitively, what you wrote makes sense. In calculus we use phrases like "vanishingly small" for intuition but really all of this is only made rigorous using the notion of a limit.

"What this essentially means" is that if $f$ is a function of $x$ and $y$, and we want to know "how much $f$ changes" at a point $(x_{0},y_{0})$, this change is going to depend on how much $x$ and $y$ change, weighted by how greatly changes in $x$ and $y$ at the point $(x_{0},y_{0}$) affect the value of $f$. The partial derivatives $f_{x}$ and $f_{y}$ tell you just that: how much the value of $f$ changes relative to changes in $x,y$ (respectively). Now, the equation you wrote encapsulates the idea, but keep in mind that the values of the partial derivatives depend on the point we're looking at.

It might help to compare this to the analogue of the formula you wrote for functions of one variable; in this case it will be $$df = f'(x)\ dx$$ To continue the above analogy, this says "changes in the value of $f$ are determined by the derivative $f'$ multiplied by the vanishingly small change in $x$". Here, $f'(x)$ is our usual derivative; for functions of multiple variables we need to instead look at partial derivatives.

There is a bigger picture here: linearization. Recall that for functions $f(x)$ of one variable, we have the formula

$$f(x)\approx f(x_{0})+f'(x_{0})(x-x_{0})$$ for $x$ near $x_{0}$. In other words, if you zoom in near the point $x=x_{0}$, the graph of $f$, which is a $2$-dimensional curve, will look like a (one-dimensional) line, and the equation of that line is the one given above, with $\approx$ replaced by $=$, $\ df$ replaced by $f(x)-f(x_{0})$ and $dx$ replaced by $(x-x_{0})$. We refer to this as the tangent line to the graph of $f$ at $x=x_{0}$.

For functions of two variables, the graph is a surface in $3$-space, but if we zoom in near a point, the graph looks like a plane (a two-dimensional object). The corresponding tangent plane approximation to the graph of $f(x,y)$ at this point is $$f(x,y)\approx f(x_{0},y_{0})+f_{x}(x-x_{0})+f_{y}(y-y_{0})$$

Notice that if we write $df$ for $f(x,y)-f(x_{0},y_{0})$, $dx$ for $(x-x_{0})$ and $dy$ for $(y-y_{0})$, we get exactly the equation in your question.

This idea works in higher dimensions but things get harder to visualize!