Why do we use the $\mathrm dy/\mathrm dx$ definition when talking about rates and not $\Delta y/\Delta x$

Question

I am a student learning rates of change. Why is it always that people use the $\,dy/\,dx$ to represent a rate of change and not the definition of the slope which is $\Delta y/ \Delta x$

I think using $\,dy/\,dx$ is kind of like a cheat to use calculus and chain rules to solve those questions. But it doesn't make any difference when we want to represent the same rate as $\Delta y/ \Delta x$. As they both are ratios, there really should not be any problem when it comes to using the standard slope equation.

For example in a problem, the side of a square is increasing at the rate of $4$ meters per second. Now, if we were to represent this in the form $\,dy/\,dx$, we can manipulate this in many ways. But if we do use $\Delta y/ \Delta x$ we cannot. So is the reason we use $\,dy/\,dx$ is just so that we can exploit calculus rules or is there some other reason?

Answer 1

Suppose that $y=f(x)$ is the graph of a function $f$ that can be differentiated. If you pick two points $(x,f(x))$ and $(x+\Delta x,f(x+\Delta x))$ that lie on the curve, then the quotient $$ \frac{\Delta y}{\Delta x}=\frac{f(x+\Delta x)-f(x)}{\Delta x} $$ represents the average rate of change between two points. On the other hand, $\frac{dy}{dx}$ is defined as $$ \frac{dy}{dx}=\lim_{\Delta x \to 0}\frac{\Delta y}{\Delta x}=\lim_{h \to 0}\frac{f(x+h)-f(x)}{h} \, , $$ and it represents the instantaneous rate of change. From this definition, it should be clear that $\frac{dy}{dx}$ is not a ratio—rather, it is the limit of the ratio $\frac{\Delta y}{\Delta x}$ as $\Delta x \to 0$. It's true that $\frac{dy}{dx}$ sometimes behaves as if it were a ratio, for instance in the case of the chain rule, but that doesn't mean that it is one. See this question for more details.

Answer 2

The thing is, $\newcommand{\dydx}{\dfrac{\textrm dy}{\textrm dx}}\dydx$ is not just a ratio of two quantities $\textrm dy$ and $\textrm dx$, it essentially describes how the variable $y$ changes with $x$.

This is one of the misconceptions beginners in calculus often have. $\dydx$ is, in no way a ratio; it beautifully converts your average rate $\dfrac{\Delta y}{\Delta x}$ into a rate which can be used to find the exact magnitude of rate at any $x$ you want.

You can observe it geometrically as well. For any arbitrary continuous graph, take two points on it and join them. The slope of the line that you get is your average rate between the two points. What the derivative does is describe the slope when these two points come closer and closer, close than any real number.

The average rate over an interval describes the overall rate, so you cannot know the rate at some point between the interval. But with instantaneous rate, you describe that at every point and can even take the average of the rates at endpoints to get your average rate back! For example, consider $y=x^2$. Let's calculate the average rate between $a$ and $a+3$. $$r_{\text{avg}}=\dfrac{(a+3)^2-a^2}{a+3-a}=2a+3$$ Now, we don't know here, what the rate is at $x=a+1$. But with derivatives, we have $$\dydx=2x$$ And thus we know the rate at $a+1$, i.e. $$r_{x=a+1}=2a+2$$ And between $a$ and $a+3$ is $$r_{a\text{ to }a+3}=\dfrac{1}{2}\Bigg(\left(\dydx\right)_{x=a+3}+\left(\dydx\right)_{x=a}\Bigg)$$ $$=\dfrac12\Big(2(a+3)+2a\Big)=2a+3$$ as expected.

Hope this helps. Ask anything if not clear :)

Answer 3

As others have elaborated, $\frac{dy}{dx}$ is not a ratio, but a rate, and $\frac{dy}{dx}$ is the limit of $\frac{\Delta y}{\Delta x} = \frac{f(x+t) - f(x)}{x - t}$ as $t$ approaches 0. This is a very good foundation to build, but personally I didn't fully understand this definition until I had a few math classes above Calculus under my belt. So I'd like to present a less rigorous but hopefully more intuitive argument.

$\frac{\Delta y}{\Delta x}$ is a generally a specific value for some pair of $x$'s(or $y$'s). For example, what is the difference in $y$ from $x =0$ to $x=1$ for some function? Let's look at four function

$$y = black(x) = x$$ $$y = red(x) = x^2$$ $$y = green(x) = \sqrt{x}$$ $$y = blue(x) = \sin(x * \frac{\pi}{2})$$

If we look from 0 to 1, each of these has $\frac{\Delta y}{\Delta x} = \frac{y(1) - y(0)}{1 - 0} = 1$. However, if we plot each of these we see that each of them are very different functions, even from 0 to 1.

While the ratio of change in $y$ vs change in $x$ is useful, it does not provide a complete picture of the rate of change of a function. The derivative allows us to define the rate of change in general, and also at any particular point $x$ for some function of $x$. This is of considerable importance among a variety of fields as your functions become more and more complex.