When talking about limit subject, all the time we point functions. But in derivative definition we use $$\frac{\Delta Y}{\Delta X}$$. Is $$\frac{\Delta Y}{\Delta X}$$ a kind of function? I have this question about integral formula as well.

• $h\mapsto\frac{f(x+h)-f(x)}{h}$ is a function, and the derivative of a function $f$ at $x$ is the limit of this function as $h\to0$ May 28 at 13:33
• When we use $\frac{\Delta y}{\Delta x}$ it is understood that $y=f(x)$. May 28 at 14:01
• @FShrike Thanks alot. what about this: $\Sigma f(x)dx$ for Integrals? Is this a function also? May 28 at 14:05

Yes, $$\Delta Y/\Delta X$$ is a kind of limit. So is the integral. But there are some nuances that can make this confusing.

$$\Delta Y/\Delta X$$ is an older, traditional notation. This can be helpful but also confusing. That's why many modern treatments use something like $$\frac{f(x+h)-f(x)}{h}$$ instead. Here, $$f$$ is the function $$X\mapsto Y$$ (so $$Y=f(X)$$), $$h=\Delta X$$, and $$\Delta Y=f(x+h)-f(x)$$. The newer notation makes explicit the function $$f$$. It also gives us the expression $$(f(x+h)-f(x))/h$$ to parse.

Strictly speaking, $$(f(x+h)-f(x))/h$$ is a function of two variables: $$x$$ and $$h$$. We could write it $$D(x,h)=(f(x+h)-f(x))/h$$. It is undefined when $$h=0$$. If we pick a particular value for $$x$$ and let $$h$$ vary, then it becomes a function of just one variable, $$h$$. People often use the term parameter for something like $$x$$: a variable whose value we fix, to focus attention on the remaining variables.

Since $$(f(x+h)-f(x))/h$$ is a function of $$h$$, but undefined at $$h=0$$, we can now try to take the limit $$\lim_{h\to 0}(f(x+h)-f(x))/h$$. If the limit exists, then $$f$$ is differentiable for that value of $$x$$. Here, $$h$$ is a so-called dummy or bound variable in the limit expression: the limit, if it exists, does not depend on $$h$$, but only the function $$f$$ and the fixed value $$x$$. "Unfixing" the value $$x$$, we then have a function of $$x$$, namely $$f'(x)$$.

In a sense, $$f$$ itself is a variable in the difference quotient. You can't evaluate $$(f(x+h)-f(x))/h$$ until you know what your function $$f$$ is. Of course, it's a different kind of variable: not a real number, but a function from real numbers to real numbers. So we can even consider the function $$f\mapsto f'$$, the "derivative function", whose domain is the collection of all differentiable functions.

With that in mind, look at the integral $$\int_a^b f(x) dx$$ This is a function of $$a$$, $$b$$, and $$f$$. The variable $$x$$ is again a bound or dummy variable: the integral does not depend on it. In more advanced treatments, you'll often see just $$\int f$$.

If we fix $$a$$ and $$f$$ and just let $$b$$ vary, we now have a function of one variable, and the famous fundamental theorem of calculus (in one form) says that the derivative of this function is $$f$$. We can also study $$f\mapsto \int f$$ as a "function of functions".

The definition of the integral $$\int_a^b f(x) dx$$ as a limit is considerably more complicated than the definition of the derivative as a limit. The simplest version is the Riemann integral with evenly spaced partitions, and sample points at the right: $$\int_a^b f(x)dx = \lim_{n\to\infty}\sum_{i=1}^n f(x_i)\Delta x$$ where $$\Delta x=\frac{b-a}{n}$$ and $$x_i = a+i\Delta x$$. Inside the summation symbol we have an expression which is a function of $$f$$, $$x_i$$ and $$\Delta x$$. But the definitions of these in terms of $$a$$, $$b$$, $$i$$, and $$n$$, makes this a function of $$f$$, $$a$$, $$b$$, $$i$$, and $$n$$. Summing on $$i$$ turns $$i$$ into a dummy variable. Taking the limit as $$n\to\infty$$ leaves us with a function of $$f$$, $$a$$, and $$b$$, the definite integral.

• Thanks. Is $\Sigma f(x)dx$ which use in integral definition also a kind of function? May 28 at 14:12
• @Russel Riemann integrals are defined as / can be defined as limits of a sequence of real numbers. These are functions $\Bbb N\to\Bbb R$, so, if you like. May 28 at 14:13
• @FShrike Thanks a million. Now I know where I am in calculus. So we use functions all the time we are in derivitive or integral definitions. May 28 at 14:21
• @Russel As long as you know precisely how the limit is defined it doesn’t really matter whether or not you think about things as limits of a sequence of numbers or as limits of a function. If you want to be super formal, most things in maths are encoded with functions and sets anyway May 28 at 14:29
• I've added some more about the integral. May 28 at 14:34