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After rewriting the definition of derivatives (which is the limit definition), we still treat $\frac{dy}{dx}$ as a fraction, for example, $\int {sin(x)cos(x)dx}$ , we substitute that $cos(x) = \frac{dy}{dx}$ and $sin(x) = y$ , then we cancel $dx$ with $dx$ as it's a fraction.

In fact, after rewriting the definition of derivatives as $\frac{df}{dx}= \grave f(x)= \lim\limits_{h \to 0} \frac{f(x+h)-f(x)}{h}$ , it's no more a fraction , but we still treat it as a fraction like the example above.

So, How can both be correct treating it as a fraction and the limit definition, How can we reconcile between treating it as a fraction and it's not actually a fraction by the limit definition?

Martin Sleziak
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Mohammad Alshareef
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    In the integral what you are really doing is using that $(f(g(x)))'=f'(g(x))g'(x)$, the chain rule, which implies that $\int f'(g(x))g'(x) =\int (f(g(x)))'=f(g(x))=f(y)= \int f'(y)$ for $y=g(x)$. – conditionalMethod Oct 26 '19 at 13:58
  • Relevant: https://math.stackexchange.com/questions/21199/is-frac-textrmdy-textrmdx-not-a-ratio/21209#21209 – JavaMan Oct 26 '19 at 14:02
  • But don't completely believe anyone telling you that it is not a fraction. There are ways to define what are $df$ and $dx$ at each point. In single real variable they turn out to be multiples of each other. You can take their quotient $df/dx$, which will be a function of the point in question. Then, finally prove that that function equals the function that you defined with the limit. So in the equality $\frac{df}{dx}=f'$, the left hand side is a fraction, but the equality has to be proven. – conditionalMethod Oct 26 '19 at 14:12
  • @ConditionalMethod: That depends on some identification of co-vector fields with their coefficients, which again depends on some choice of basis. I'm just saying these statements are only actually true up to an interpretation of one sort or another. – WoolierThanThou Oct 26 '19 at 16:10
  • @conditionalMethod, What are f(x) and g(x) in the integral ?, I still haven't got it. – Mohammad Alshareef Oct 26 '19 at 16:45
  • In your question $f(x)=x$ and $g(x)=sin(x)$. – conditionalMethod Oct 26 '19 at 17:00
  • @conditionalMethod, Sorry, But do you mean that $f'(x) = x$ or $f(x) = (x^2)/2$, could you please check it. – Mohammad Alshareef Oct 26 '19 at 17:51
  • Yes, $f'(x)=x$. – conditionalMethod Oct 27 '19 at 00:46

2 Answers2

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It's not a fraction. What we have here is an instance of suggestive notation!

The symbol you're using looks like a fraction, and if you try to treat it like a fraction, you develop naïve results that turn out to also match actual results. This, to me, is a sign that the notation is well-chosen.

What actually happens is an application of the fundamental theorem of calculus. Since $\frac{d}{dx} f(g(x))=\frac{d}{du}f|_{u=g(x)}\cdot\frac{d}{dx} g(x)$ for differentiable functions $f$ and $g$, you can just check that $\int_a^b f(g(x))g'(x)\textrm{d}x=\int_{g(a)}^{g(b)} f(u)\textrm{d}u$. Note that the placeholder variables $x$ and $u$ don't carry inherent meaning.

It just happens that this actual mathematical result is consistent with what would happen if you just did your naïve fractional calculus.

WoolierThanThou
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  • How can you also explain other properties like: $\frac{dx}{dy} = \frac {1}{\frac{dy}{dx}}$ and so on. – Mohammad Alshareef Oct 26 '19 at 14:19
  • What sort of explanation are you looking for? The answer is the same: It's true for fractions, but that doesn't prove that it's true for derivatives. However, the fractional notation might give you the idea that such a statement is true. And... as it turns out, you can prove it using the definition of the derivative - this is the inverse function theorem in one dimension. – WoolierThanThou Oct 26 '19 at 14:23
  • I am sorry, I am really confused, just one more, when we say that $dy=\frac{dy}{dx} dx$ what do we mean by $dy$? – Mohammad Alshareef Oct 26 '19 at 15:02
  • Depends on which sort of mathematics you're doing. In classic calculus, the answer is "nothing really, but the notation is convenient". In measure/probability theory, $dy$ and $dx$ are measures, while $dy/dx$ is a density function, and the statement is still true. In differential geometry, $dy$ and $dx$ are differential forms and $dy/dx$ is a smooth function and again, the statement is true. – WoolierThanThou Oct 26 '19 at 15:05
  • What do you mean by nothing, Can I just ignore them or delete them from the equation!!? (I MEAN IN CALCULUS) – Mohammad Alshareef Oct 26 '19 at 15:50
  • I mean that your equation doesn't actually mean anything in calculus. It is, however, useful notation for recalling how change of variables work. I'm saying that your formula in calculus doesn't mean anything as far as rigorous mathematics go. For instance, while your formula is in some sense "true" - i.e. change of variables actually does work - you cannot prove that your above formula is true by saying that "dx cancels with 1/dx". – WoolierThanThou Oct 26 '19 at 16:08
  • Got it, Thank you very much – Mohammad Alshareef Oct 26 '19 at 16:12
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The symbol $$\int f(x)dx $$ simply means the integral is with respect to $x$ and you may even drop $dx$ if there is only one variable involved.

Mohammad Riazi-Kermani
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