Does delta x cancel with dx in Euler's method?

So I am a High School student currently learning about Euler's method. The formula we were given in class was:

$$y_1 = \frac{dy}{dx}(x_0, y_0) \cdot \Delta x + y_0.$$

The only thing I don't really get about the formula, is the $$\Delta$$x part. Does it cancel out with the dx? I know dx is technically a differential, but with Euler's method you are essentially 'extending' the tangent ($$\frac{dy}{dx}$$) out until it has travelled $$\Delta$$x in the $$x$$ direction. In this way, does the $$dx$$ cancel with the $$\Delta x$$?

Because if it did cancel, then the formula would really make sense: The y value at some point equals the change in $$y$$ due to the derivative being applied over the $$\Delta x$$ interval, plus the initial $$y$$ value.

• No, besides the fact that they are different objects, the heart of Euler's method is to approximate the original function $y(x)$ by its linear approximation $y(x_0) + y'(x_0)(x - x_0)$ on a small interval $[x_0, x_0 + \Delta x]$, and so, they are in general different. – Sangchul Lee Mar 21 at 1:32
• @SangchulLee hmm, so why is delta x multiplied to the derivative at the starting point? Why not divide? Or some other operation? – Addison Mar 21 at 1:33
• The derivative at the initial point is just some number, say $m$, that is equal to the slope of the tangent line to the curve at that point. In Euler's method, we move to the next $y$-value by moving along this tangent line (with slope $m$). If you call the new point $y_1$, then since the slope of the line is $$\color{blue}{\frac{y_1-y_0}{\Delta x}=m},$$ we find that $y_1 = y_0 + m\Delta x$. So the multiplication comes about by rearranging the equation for a slope. – Minus One-Twelfth Mar 21 at 1:43
• @MinusOne-Twelfth alright, fair enough. Thank you! – Addison Mar 21 at 1:45
• Good for you for taking the care to format the formulas in MathJax until they were readable. It shows consideration for your reader and good attention to detail. In the future, you can do it even better by putting each formula completely inside its own set of $...$ delimiters, that is, go into MathJax, write the whole formula, and get out of MathJax when you have to write the next thing in English, rather than popping in and out of MathJax several times in a single equation. – David K Mar 21 at 2:36

Technically, the $$dx$$ in $$\frac{dy}{dx}$$ cannot cancel anything, because the symbol $$\frac{dy}{dx}$$ in $$\frac{dy}{dx}(x_0,y_0)$$ is the name of a single function that takes the two numbers $$x_0$$ and $$y_0$$ as input and produces a third number as output. It isn't anything divided by anything, least of all something called $$dy$$ divided by something called $$dx,$$ at least in the standard modern version of calculus.
I say "standard modern" because Leibniz may have thought of $$\frac{dy}{dx}$$ as a ratio when he invented the notation in the 17th century, and because the derivative might be considered a ratio in non-standard analysis, which I'm not sufficiently familiar with. What I mean by "standard modern" is the approach that dominated calculus in the 20th century and is still used in typical textbooks today.
But I think the reason the notation $$\frac{dy}{dx}$$ held on for so long despite the "standard modern" interpretation (which made nonsense of writing it as a ratio) is that it has great mnemonic power. You can look at something like $$\frac{dy}{dx}(x_0, y_0) \cdot \Delta x$$ and see that it is properly formed; the $$dx$$ does not literally cancel the $$\Delta x$$ mathematically, but it feels like it is doing so in a vague intutive way. You would realize instantly that something was wrong if you saw $$\frac{dy}{dx}(x_0, y_0) \cdot \Delta y,$$ because $$dx$$ and $$\Delta y$$ don't "cancel" in that vague intuitive way.