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If you think about the limit definition of the derivative, $dy$ represents $$\lim_{h\rightarrow 0}\dfrac {f(x+h)-f(x)}{h}$$, and $dx$ represents

$$\lim_{h\rightarrow 0}$$ . So you have a $\;\;$$\dfrac {number}{another\; number}=a fraction$, so why can't you treat it as one? Thanks! (by the way if possible please keep the answers at a calc AB level)


marked as duplicate by mrf, Ayman Hourieh, MathOverview, draks ..., Michael Albanese Jun 3 '13 at 10:41

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  • $\begingroup$ What does "and $dx$ represents lim as $h$ approaches $0$ of $h$" mean? Are you saying that $dx = 0$? In that case you have answered your own question. If not, you need to explain what you mean (and possibly reinvent non-standard analysis as you do). $\endgroup$ – mrf Jun 3 '13 at 9:28
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    $\begingroup$ You should read the answers to math.stackexchange.com/questions/21199/is-dy-dx-not-a-ratio $\endgroup$ – mrf Jun 3 '13 at 9:30
  • $\begingroup$ Thanks, that link really helped $\endgroup$ – Ovi Jun 3 '13 at 9:36
  • $\begingroup$ math.stackexchange.com/questions/1548487/… $\endgroup$ – user117644 Dec 26 '15 at 22:58

The derivative, when it exists, is a real number (I'm restricting here to real values functions only for simplicity). Not every real number is a fraction (i.e., $\pi$ is not a fraction), but every real number is trivially a quotient of two real numbers (namely, $x=\frac{x}{1}$). So, in which sense is the derivative a fraction? answer: it's not. And now, in which sense is the derivative a quotient to two numbers? Ahhh, let's try to answer that then: By definition $f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$. Well, that is not a quotient of two numbers, but rather it is a limit. A limit, when it exists, is a number. This particular limit is a limit of quotients of a particular form (still, not of fractions in general, but quotients of real numbers).

The meaning of the derivative $f'(x)$ is the instantaneous rate of change of the value of $f$ at the point $x$. It is defined as a certain limit. If you now intuitively think of $h$ as an infinitesimal number (warning: infinitesimals do not exist in $\mathbb R$, but they exist in certain extensions of the reals) then you can consider the single expression $\frac{f(x+h)-f(x)}{h}$. In systems where infinitesimals really do exist one can show that this single expression, when the derivative exists, is actually infinitesimally close to the actural derivative $f'(x)$. That is, when $h\ne 0$ is infinitesimal, $f'(x)-\frac{f(x+h)-f(x)}{h}$ is itself infinitesimal. One can them compute with this expression as if it were the derivative (with some care). This can be done informally, and to some extend this is how the creators of early calculus (prior to Cauchy) argued, or it can be done rigorously using any one of a number of different techniques to introduce infinitesimals into calculus. However, getting infinitesimals into the picture comes with a price. There are logical/set-theoretical issues with such models rendering all of them not very explicit.

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    $\begingroup$ Nice reasoning and very well explained! $\endgroup$ – Rohinb97 Jun 3 '13 at 10:40

Practically speaking you can pretty much just treat derivatives as fractions but formally speaking we like to say that $\frac{\mathrm{d}y}{\mathrm{d}x}$ is one object. It's not just a small change in $y$ over a small change in $x$, it's the small change in $y$ when we make a small change in $x$.

Keep in mind, $\mathrm{d}y$ and $\mathrm{d}x$ aren't really numbers either, they're infinitesimals. It's only the "ratio" of them that comes out to be a number.


The alternative approach to solve the problem \begin{eqnarray} y'(x) & = & \alpha y(x) \\ y(0) & = & 1 \end{eqnarray} is to find the general solution of the homogenous equation and make manipulations \begin{eqnarray} \frac{y'(x)}{y(x)} & = & \alpha \\ \ln(|y(x)|) & = & \alpha x + C \\ |y(x)| & = & e^{\alpha x + C} = e^C e^{\alpha x} = D e^{\alpha x}, \ \ D > 0 \\ y(x) & = & D e^{\alpha x} \ , \ \ \ D \neq 0 \ . \end{eqnarray} The special solution of the homogenous equation is found by inserting $D = 0$, that doesn't belong to the set of solutions found above. Hence we have \begin{eqnarray} y(x) = D e^{\alpha x} \ . \end{eqnarray} Now using the initial condition we obtain \begin{equation} 1 = y(0) = D e^{\alpha 0} = D \ . \end{equation} Hence \begin{equation} y(x) = e^{\alpha x} \ , \end{equation} that is the solution of the problem.

However the method of separation of variables followed by integration is an efficient way to solve differential equations and it is the original way. The problem with it is the separation phase of the solution process where both sides of the equation reduce to the form $0=0$ on the limit where infinitesimals $dx$ and $dy$ approach to zero. In the above solution process all phases stay on the limit where the derivative is defined. If interpreted to the infinitesimal form it follows to a limit of the fraction of $dy$ and $dx$ that approaches to the nonzero derivative.

Yet another aspect to the question. The fraction $\frac{dy}{dx}$ is a historical notation for derivative from the age when there wasn't yet proper definition of limit. This motivates a representation theorem for derivative. We recall the definition of derivative and calculate \begin{eqnarray} y'(x_1) & = & \lim_{h \rightarrow 0} \frac{y(x_1+h)-y(x_1)}{h} = \lim_{x_1+h \rightarrow x_1} \frac{y(x_1+h)-y(x_1)}{x_1+h-x_1} = \lim_{x_2 \rightarrow x_1} \frac{y(x_2)-y(x_1)}{x_2-x_1} \ , \end{eqnarray} where we have set $x_2 = x_1 + h$. Hence we have the representation \begin{equation} y'(x_1) = \lim_{x_2 \rightarrow x_1} \frac{y(x_2)-y(x_1)}{x_2-x_1} \ . \end{equation} For small differentials $dx = x_2 - x_1$ and $dy = y(x_2) - y(x_1)$ we have \begin{eqnarray} \frac{dy}{dx} & = & \frac{y(x_2)-y(x_1)}{x_2-x_1} \ . \end{eqnarray} We can call $dx$ and $dy$ differentials if and only if $x_2 \neq x_1$ because $x_1$ and $x_2$ have to differ. Otherwise $dx=0$ and we don't want to divide by $0$. It would contradict the field axioms. At $x_1$ we have \begin{equation} \frac{dy}{dx} \rightarrow y'(x_1) \end{equation} Similarly at $x$ we have \begin{equation} \frac{dy}{dx} \rightarrow y'(x) \ . \end{equation} This means that we should interpret $\frac{dy}{dx}$ as a difference quotient that has at $x$ the limit $y'(x)$. Hence also the notation \begin{eqnarray} \frac{dy}{dx} & = & \alpha x \\ y(0) & = & 1 \end{eqnarray} contains a weakness, but defines an equation on the limit that has a unique solution. Hence there is nothing wrong with treating $\frac{dy}{dx}$ as a fraction, but then it is not the definition of derivative.

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    $\begingroup$ In what way is this an answer to the question? $\endgroup$ – mrf Jun 3 '13 at 10:41
  • $\begingroup$ This method is an example of the use of backwards differentiation of composite function instead of separation of variables that is of the form $0=0$ on the limit, where $dx$ and $dy$ approach to $0$. $\endgroup$ – Juha-Matti Vihtanen Jun 3 '13 at 12:11

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