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Say we have $\frac{dy}{dx}=e^x$, then it is solved by taking $dx$ to right hand side of the equation and integrated.

My question is if $\frac{d}{dx}$ is an operator, then how do can take $dx$ to right? What is the explanation and reason behind this mathematically and intuitively?

Student
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  • Try to think of $\frac {dy}{dx}$ as a fraction: Let $dy$ denote a small change in $x$ and $dx$ denote a small change in $x$. By dividing, you get the gradient. – For the love of maths Jan 28 '18 at 08:35
  • $$y = \int \dfrac{dy}{dx} dx = \int e^x dx = e^x + C$$ – user8277998 Jan 28 '18 at 08:42
  • $dy$ and $dx$ are two differentials and $\frac{dy}{dx}$ is a derivative. Thinks about the relation between differential and derivative, especially in the multivariate calculus, and you will see. – Zongxiang Yi Jan 28 '18 at 08:52
  • Just as $dy/dx$ is not a fraction but some operation applied on $y$ the symbol $\int y, dx$ is not a sum of things like $y$ multiplied by $dx$. Rather it is another operation applied on $y$ and unlike $dy/dx$ it does not give unique answer rather a family of answers all differing by constant. And these operations are sort of inverses to each other. – Paramanand Singh Jan 28 '18 at 08:56
  • Differentiation is an operator, not a quotient. Integration is an operator, not a sum. Choosing notation that makes it look like differentiation is a quotient is common, but unfortunate. It leads to this question being asked. The fundamental theorem of calculus relates these two operators and that is what really goes on in your calculation. – Ittay Weiss Jan 28 '18 at 09:28

3 Answers3

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In general, for a function of $n$ variables $y(x_1,\dots,x_n)$, its differential is

$$dy = \frac{\partial y}{\partial x_1}dx_1+\dots+\frac{\partial y}{\partial x_n}dx_n$$

In case $n=1$, we obtain (saying just $x$ instead of $x_1$):

$$dy = \frac{\partial y}{\partial x}dx$$

Which is commonly abused like this:

$$\frac{dy}{dx} = \frac{\partial y}{\partial x}$$

Note that in general dividing differentials makes no sense, much like dividing vectors in $n$-dimensional vector space. But your somewhat can divide them in 1-dimensional space, and that's exactly what is happening here with the differentials.

lisyarus
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It is easier to understand if you look at taking $dx$ to the right side and integrating as one operation. In fact, integration does not make any sense if no differential is present.

This way you are just applying an operator to both sides.

$$\int \frac{dy}{dx}dx=y+C$$

$$\int e^xdx=e^x+C$$

Sergey Guminov
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Don't think too far. This is just a brainless, but easy to memorize step in a solution recipe. Of course it can be proven (and most probably you have seen a proof) that the end result obtained in this way is correct.

Check the answers to the question quoted in Hans Lundmark's comment.

Christian Blatter
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