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I'd like to know the reason behind some notations used when handling derivatives and integrals.

For example, why does $x' = 1 \frac{d}{dx}$ and not simply $1$? Related to this, why is integral of that written as $\int 1dx$ and not $\int 1 \frac{d}{dx}$?

Also, when writing out the result of an integral why should we add the constant? as in $\int x dx = \frac{x^2}{2} + C$.

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First, the derivative can be rewritten as a difference quotient. Let's do it step by step.

Let $f(x)=x$ be a function and derive it by its only dependent $x$. The derivative $\frac{d}{dx}f(x)$ of a function $f(x)$ can be interpreted as the slope of the function at a point $x$. Let's do that visually by drawing the graph of the function $f(x)=2x+1$. Now draw a little triangle below (or above) the graph between $x_1=1$ and $x_2=2$, thus between $f(x_1)=3$ and $f(x_2)=5$.

Now use this triangle to calculate the slope of $f(x)$, which gives you the difference quotient:$$\frac{f(x_2)-f(x_1)}{x_2-x_1}$$ But we do not want the averaged slope between $x_1$ and $x_2$, but the slope in a single point $x$.

The idea is, to let the difference of $x_2$ and $x_1$ become very small (colloquial: zero). So, the very general form of a difference quotient is $$\frac{f(x+\delta x)-f(x)}{\delta x}$$ where $\delta x$ is the distance between $x_1$ and $x_2$ and thus $\delta x \rightarrow 0$.

And this difference quotient is nothing else but the derivative of $f(x)$ derived by the variable $x$. And the origin of the '$d$' in $dx$ becomes clear, as it is the $\delta$ in $\delta x$. (with partial derivatives you usually do not write '$d$' but '$\delta$')

Second, with an integral, you do it the other way round. Finding the function $f(x)$ when you've got its derivative $f'(x)$. Thus, you are not calculating a difference quotient any more and thus you do not have $\frac{d}{dx}$.

You write $dx$ at the end of an integral, to show, that the function is integrated by the variable $x$. This becomes very important, when you deal with functions, which are dependent on more than one variable. (Edit: Ha01 gave a better explanation on this.)

Third, when you derive our function $f(x)=2x+1$, it gives you $f'(x)=2$. The constant term $1$ disappears. Now take another function $g(x)=2x+5$. Its derivative $g'(x)$ is as well constant $2$.

When you calculate the anti-derivative, you have to take into account that any constant term will disappear with every derivative. And thus not to specify one single anti-derivative, you denote the constant term $C$ to describe a family of functions.

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  • $\begingroup$ K.: If you want $\Delta x$ (which is rather more common than $\delta x$) you can use \Delta to produce $\Delta$. $\endgroup$ Apr 19, 2011 at 20:31
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For the first question - it depends on the variable that you derive by. If you derive $x$ by $x$, i.e. $x'=\frac{dx}{dx}=1$ then you simply wtite 1. As to the second question, the indefinite integral should give you an "anti-derivative" of the function, and there are a lot of them, but they differ by a constant, hence you add $C$, to represent the family of constants.

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  • $\begingroup$ More specifically, $C$ represents the kernel of the derivative. Hence, in order to define an anti-derivative you need to look at the factor space, modulo the kernel. $\endgroup$ Apr 19, 2011 at 19:25
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The notation $x'$ is ambiguous. If the prime ($'$) is differentiation with respect to $x$, then $x' = 1$. Otherwise if the prime denotes differentiation with respect to some other variable, say $t$, then $x' = \frac{dx}{dt}$.

You can think of the expression $\frac{d}{dx}$ as an operator acting on functions $f(x)$ by differentiation with respect to $x$. Now, $dx$ by itself is something quite different, being really a measure (or one of several notions). In elementary calculus, it is a little object which tells us how to measure the length of intervals. It is not a derivative!

For the second question, the derivative of any constant is zero, and so to be correct one must add on a constant (which may be zero) when integrating.

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At first, $dx$ was just a notation. The reason was that integral is like a sum: $\int x dx $ ~ $ \sum_{i=1}^{n-1} x_i \cdot (x_{i+1}-x_{i})$. The expression $x_{i+1}-x_{i}$ was denoted by $\Delta x$ and so integral is a sum where $\sum$ is transformed to $\int$ and $\Delta x$ is transformed to $dx$.

But then, the theory of differential forms was created and $dx$ got an actual meaning as $d$ applied to $x$. Now it is gets complicated ;). Here $x$ is a function, $d$ is an operator that transforms differential forms of order $k$ to differential forms of order $k+1$. Functions are differentail forms of order $0$, so $dx$ is a differential form of order $1$.

The differential form of order $k$ is not a very complicated object in its simplest form. You can imagine it as a function of a point and $k$ vectors. This function must be linear with respect to each of $k$ vectors and it must change a sign if you exchange any two vector arguments (like that: $f(x,v_1,v_2,...,v_k) = -f(x,v_2,v_1,...,v_k)$).

$dx$ is defined as a function that takes a point $x$ and vector $v$ and returns $v_x$ - the coordinate of vector $v$ along the axis $x$ (so this function doesn't even depend on the point $x$!). This function linear with respect to $v$ and there is only a single vector argument and so there is nothing to exchange.

$f(x)dx$ is a function that takes a point $x$ and vector $v$ and returns $f(x)v_x$.

There is a general rule that you can take a differential form of order $k$ and a $k$-dimensional surface and calculate an integral of this form along this surface.

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    $\begingroup$ I think your history is not quite right; in fact, "at first" (i.e., with Leibnitz), "dx" had an actual, if somewhat fuzzy, meaning (and the integral sign was really just a large $S$, standing for sum). $\endgroup$ Apr 19, 2011 at 20:18
  • $\begingroup$ Leibnitz wouldn't be able to explain what this $dx$ is as an object on a modern exam unlike an integral. So I consider it as part of the notion. Mathematicians tried to deal with $dx$ as number back then and as result we have a zoo of strange notations today. $\endgroup$
    – Ha01
    Apr 19, 2011 at 20:40
  • $\begingroup$ @Hao1: While Leibnitz wouldn't be able to explain it in a modern exam (lacking the notions), that does not change the fact that the notation originates with him and that when he introduced it, "dx" did have a meaning, contrary to your assertion that "at first [it] was just notation". At first, it wasn't. Then it became "just notation" later. $\endgroup$ Apr 20, 2011 at 1:26

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