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I'm currently studying physics from a calculus perspective, and in the process I'm improving my intuition of calculus. I'm struggling with a sort of chicken and egg problem: I have several identities, but I'm not sure which ones come first, or which ones I can use to derive later ones.

When I first started learning calculus in high school, we learned derivatives using Lagrange's notation, and always referred to it as a rate of change: $$\text{The derivative of } f(x) = f'(x)$$

We then learned Leibniz's notation, but only that it was an alternative form of the prime notation (as opposed to being an equality): $$f'(x) \text{ can be written as } \frac{dy}{dx}$$

Now I'm learning calculus from a more rigorous perspective, using this as a reference. Early on, they define the differential: $$\text{The differential of } f(x) \text{ is } df = f'(x)dx$$ Which is a comfortable syntax to me. I also recall from before that (leaving out the constant of integration): $$\int{f'(x)dx} = f(x)$$

And herein lies the chicken and egg problem. Since my assumption was that Leibniz's notation can replace Lagrange's, I can write: $$\int{f'(x)dx} \text{ as } \int{\frac{dy}{dx}dx}$$ From differential equations I recall that we can cancel the dx terms as we do in Algebra: $$\int{\frac{dy}{dx}dx} = \int{dy}$$

Which matches what appears when we integrate the differential, as is done in the MIT course: $$df = f'(x)dx$$ $$Equation A: \int{df} = \int{f'(x)dx}$$ From the previous statement of the integral, we can see that on the left side of equation A (again leaving out the constant of integration): $$\int{df} = \int{df\frac{dx}{dx}} = \int{\frac{df}{dx}dx} = \int{f'(x)dx} = f(x)$$ Which we could have done in reverse. Importantly this leaves us with: $$Equation B: \int{df} = f(x)$$

Which begs my question: which form came first? My intuition was to start from the fundamental theorem of calculus (using primes instead of capitals for clarity): $$f(x) = \int{f'(x)dx}$$ Which implies that the right side of equation A is what we know, and that equation B is a corollary of it. It also implies that the integral only fundamentally makes sense in the $g(x)dx$ form, which matches with my internal understanding of integrals as: $$\lim_{\Delta{x} \to 0} \sum{g(x)\Delta{x}}$$

But a corollary of that assumption is that $\int{df}$ doesn't make sense on its own, and it's only by multiplying by $\frac{dx}{dx}$ that it can be integrated. This seems to run contrary to some learning resources I've run across, which use $$\int{df} = f(x)$$ as though it were the fundamental rule.

So to repeat my question: which comes first? Which one do we use to derive the other? Is Equation A necessary to have Equation B, or is it the other way around?

I realize I've left out the constant of integration in multiple places, but I don't believe it's relevant to my question.

Thanks in advance!

Edit:

The accepted answer to this question seems to imply that Equation B is more fundamental, and that the fundamental theorem of calculus lets you transform between that and Equation A. But the accepted answer to this question seems to imply (less strongly) that it is indeed the fundamental theorem of calculus that is more... fundamental.

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  • $\begingroup$ When two statements are both true, there is no "which came first". They are simply true. You can prove one first, then use it to prove the other, or you can prove the other first, then use it to prove the one. Or you can prove each without any reference to the other. As long as you don't try to prove both from each other (circular reasoning), everything is copacetic. $\endgroup$ – Paul Sinclair Mar 5 '17 at 21:34
  • $\begingroup$ @PaulSinclair I understand, but I think I only have circular proofs above. By "first" I really mean the axiom we started with to derive the other. So perhaps what I really need is to treat y' = dy/dx as a definition, not a derivation, from which I can use the fundamental theorem of calculus to transform between differentials and integrals. $\endgroup$ – Brandon Mar 5 '17 at 21:45
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The derivative of a function $y(x)$ is defined to be $$\lim_{h \to 0}\frac{y(x+h) - y(x)}h$$ $y'$ and $\frac {dy}{dx}$ are just two notations used for this limit. Neither is defined from the other. I have a name and a social security number. These are tied together legally, but neither is defined by the other. Instead, they both mean me. In this same way $y'$ is not derived from $\frac {dy}{dx}$, nor is $\frac {dy}{dx}$ derived from $y'$. Instead, they both mean that limit. A crucial point, though, is that $\frac{dy}{dx}$ is not a true quotient of two things. Instead, it is the limit of a quotient.

The definition of differential you gave, $df = f'(x)dx$, is not fully rigorous. (What is $dx$? What is the difference between independent and dependent variables? How is it that independent variables can become dependent? What is a variable in the first place?) There are ways (many different ways) it can be made rigorous, but these involve details that are just not important to novices. At your stage, it is the concept that you need to understand, not the niggling details of a rigorous development. So we just give the definition above even though it cannot stand up to great scrutiny.

So do not expect that any treatment of differentials in this way will be completely rigorous.

From differential equations I recall that we can cancel the dx terms as we do in Algebra: $$\int \frac{dy}{dx}\:dx=\int\:dy$$

This is the definition of $\int\:dy$ (applied to the special case of an integrand of $1$. It is not a theorem or a cancellation (remember, $\frac {dy}{dx}$ is not really a quotient, so we cannot truly cancel out the $dx$). It is how we define the notation for dependent variables. And we define it that way because it is suggestive of the fundamental theorem of calculus (which itself is proven for independent variables).

Your Equation A is just this same equation using $f$ instead of $y$, and writing $f'(x)$ instead of $\frac {dy}{dx}$. But what letter you use for a variable is unimportant, and the Lagrange and Leibnitz notations mean the exact same thing. So really, your Equation A is just the definition above.

On the other hand, equation B, $$\int\:df = f(x) + C$$ or $$\int\:dy = y + C$$ IS the fundamental theorem of calculus (in indeterminant form). Because equation A is the definition of $\int\:df$, equation B is just a way of writing $$\int f'(x)\:dx = f(x) + C$$.

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  • $\begingroup$ Thank you! That was very thorough and descriptive, and you answered all of my questions. $\endgroup$ – Brandon Mar 6 '17 at 0:42

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