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My question is if it is okay / mathematically rigorous to write the Chain Rule like that (the Leibniz way). I thought that $dx$, etc. do not follow the rules of algebra and cannot be treated as such. For example, I write $\int 1\, dx$, rather than $\int 1 \,dx$, and I write $\int \frac{dx}{a}$ instead of $\int \frac{1}{a}\, dx$.

So, is it correct to say $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$ and, in essence, have the $\frac{du}{du}$ cancel to $1$?

(If my notion of $dx$ is not correct, I would also like an explanation of what really that is)

edit: Is it the same deal with $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$?

Michael Hardy
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MT_
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  • With the standard meaning, it's wrong. – Git Gud Mar 19 '14 at 23:43
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    Treating differentials like fractions is kind of like how a classical mathematician would look on a calculation involving the root of a negative number. Clearly you get the right answer every time, but there's something awfully wrong with how you get there. – Arthur Mar 19 '14 at 23:45
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    Leibniz notation is great and can help you keep track of the chain rule easily but every calculus instructor should explicitly state that it is only notation and that $\dfrac{dy}{dx}$ is not a fraction but only a notation. It is very suggestive notation, but that is all it is. Neither $dy$ nor $dx$ exist as independent entities. You can ONLY take the entire thing together to make meaningful notation. I feel like Leibniz notation should be stricken from first year calculus texts for this reason, despite how useful it is.. – Cameron Williams Mar 19 '14 at 23:50
  • I dislike this notation because two different functions, $y$ and $y \circ u$, are both being denoted by $y$. Moreover, the Leibniz notation doesn't make it clear that $y'$ is evaluated at $u(x)$. – littleO Mar 20 '14 at 01:35

4 Answers4

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In terms of differentials (in the single-variable case), $\frac{dy}{dx}$ is the unique scalar with the property that $\frac{dy}{dx}dx = dy$.

$\frac{dy}{du} \frac{du}{dx}$ therefore has the property that

$$\frac{dy}{du} \frac{du}{dx} dx = \frac{dy}{du} du = dy $$

therefore $\frac{dy}{du}\frac{du}{dx} = \frac{dy}{dx}$.

You can't really justify the result by rearranging the expression like you would with fractions (e.g. by combining them into a single 'fraction'), which is why people mean when they say things like "you can't just cancel them". However, you can still prove (again this only makes sense in the single-variable case) that rearrangements are equal: e.g.

$$ \frac{dw}{dx} \frac{dy}{dz} = \frac{dw}{dz} \frac{dy}{dx} $$

(note that you could use this identity to prove your identity, because $\frac{du}{du} = 1$)

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The notation $\frac{dy}{dx} = \frac{dy}{du} * \frac{du}{dx}$ is valid. However, you cannot prove the chain rule just by "cancelling" the two $du$'s; it doesn't work that way.

Tony
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  • "it doesn't work that way" -- care to elaborate? – MT_ Mar 19 '14 at 23:48
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    No, the notation makes no sense. – Git Gud Mar 19 '14 at 23:48
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    @GitGud See Wolfram MathWorld or Wikipedia - both use this notation. In the end, arguing about notation is unimportant, and notation is defined by what people commonly use. It's just a language used to express math. – Tony Mar 19 '14 at 23:50
  • @user92774: When you write $\frac{dy}{dx}$, you aren't really writing a fraction consisting of $dy$ divided by $dx$. It's just a notation defined as the limit of the difference quotient as $x$ approaches a certain point. – Tony Mar 19 '14 at 23:53
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    No, the notation is defined, period. It has nothing to do with what people use. When one writes $f=gh$, what one means is $f(a)=g(a)h(a)$, for all $a$ somewhere. This is how it is. Particularizing to $\dfrac{dy}{dx} = \dfrac{dy}{du} \dfrac{du}{dx}$ gives $\dfrac{dy}{dx}(a) = \dfrac{dy}{du}(a) \dfrac{du}{dx}(a)$ and that's not at all what's intended. – Git Gud Mar 19 '14 at 23:55
  • Are you suggesting that Wolfram MathWorld is wrong? – Tony Mar 19 '14 at 23:57
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    @Tony They are abusing the notation and abuses of notation are wrong. Concluding: yes, they are wrong. – Git Gud Mar 19 '14 at 23:58
  • @Git Gud: Then I hope you don't use both of $sin^2$ and $sin^{-1}$. – user2345215 Mar 20 '14 at 00:11
  • @user2345215 I use neither, but $\sin ^{-1}$ has no problems. – Git Gud Mar 20 '14 at 00:12
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There is a $C$ which looks like $A/B$ (suppose) in some german notation. And now in some rule you denote some value by $C$ and others by $A$ and $B$ and $C = A/B$ is the rule. You won't say you broke the notation $C$ into $A/B$ because it was looking so.

You will deal more such cases in integration.

Abhishek Verma
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The relation $$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$$

requires that $u$ has a nonzero relationship with $x$ and $y$. This doesn't have to be explicitly stated when the chain rule is written as $$D_x f(g(x)) = f'g(x) \cdot g'(x)$$

Example:

Consider using $g = \text{gravitational force}$ and $r = \text{distance}$. You might know:

$$g = G\frac{m_1m_2}{r^2}$$

What if $u$ is your temperature? Perhaps your temperature does not change based on your location. $$\frac{du}{dr} = 0$$ $$\frac{dg}{du} = 0$$

If you applied the chain rule without realizing that $u$ is an independent variable, you'd get:

$$\begin{align}\frac{dg}{dr} &= \frac{dg}{du} \cdot \frac{du}{dr}\\ &= 0 \cdot 0 \end{align}$$

...and I think we can agree that gravitational force is not universally constant. So when using the differential form of the chain rule, make sure you are not using an independent variable.

DanielV
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