# "Undo" Second Derivative With Square Root?

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• ryaamaak
In summary: Why did it work? In summary, the professor did a bit of algebraic wizardry in a derivation for one of Kepler's Laws where a second derivative was simplified to a first derivative by taking the square root of both sides of the relation. It basically went something like this: \frac{d^2 u}{dx^2} = \big( f(x)\big)^2\\ \frac{d u}{dx} \frac{d u}{dx} = \big( f(x)\big)^2 \\\frac{d u}{dx} = f(x) .Why could my professor do this? How does this make sense in a physical sense? Or
ryaamaak
In my classical mechanics course, the professor did a bit of algebraic wizardry in a derivation for one of Kepler's Laws where a second derivative was simplified to a first derivative by taking the square root of both sides of the relation. It basically went something like this:
$$\frac{d^2 u}{dx^2} = \big( f(x)\big)^2\\ \frac{d u}{dx} \frac{d u}{dx} = \big( f(x)\big)^2 \\\frac{d u}{dx} = f(x)$$.
Why could my professor do this? How does this make sense in a physical sense? Or in a mathematical sense? What are the cases for where this trick won't work? Any clarification is much appreciated.

It looks like he may have goofed. None of us are infallible. I have seen even extremely astute professors make an error on occasion. It looks to me like he may have blundered... And no one caught the mistake in the lecture?

ryaamaak said:
something like this
Probably something 'slightly' different. Either that or he prof went nuts. Just take any solution for the last equation (e.g. ##u = x^2 \Rightarrow f(x) = 2x ## and see that the equations above it are not satisfied.

This is an abuse of the notation, as
$$\frac{d^2u}{dx^2} \neq \frac{du}{dx} \frac{du}{dx}$$
A second derivative is not a product of first derivates.

I would guess that it is either a brain fart or a trap (to see if anyone is paying attention or confident enough to point out the mistake).

ryaamaak and sophiecentaur
The professor prefaced this derivation by saying it was one he had done himself and “was the most elegant way he could think of”, then copied it down from his lecture notes. I’m assuming that precludes casual error.

As to any students pointing out the mistake, I asked after class because I was confused, but all I got was some serious side eye from the prof and no clarification. At that point I started to mistrust my own understanding of operators. To my knowledge none of the other students thought twice about his work on the board. Thank you all for being my sanity check!

ryaamaak said:
The professor prefaced this derivation by saying it was one he had done himself and “was the most elegant way he could think of”, then copied it down from his lecture notes. I’m assuming that precludes casual error.

As to any students pointing out the mistake, I asked after class because I was confused, but all I got was some serious side eye from the prof and no clarification. At that point I started to mistrust my own understanding of operators. To my knowledge none of the other students thought twice about his work on the board. Thank you all for being my sanity check!

Give your prof. the example that BvU gave and ask him why the second derivative does not match the square of the function.

Zz.

ryaamaak
It would be interesting to see the full context of this derivation. If ##f(x) = -1/x##, then ##\frac{df}{dx} = (f(x))^2##, so maybe the derivation arose in the context of the inverse square law of gravitation. That's really the only thing I can think of that's remotely related to what OP is saying.

Edit: maybe something like ##\frac{d^2u}{dr^2} \propto \frac{k}{r^2}##, therefore ##\frac{du}{dr} \propto \frac{k}{r}##? I dunno.

TeethWhitener said:
It would be interesting to see the full context of this derivation. If ##f(x) = -1/x##, then ##\frac{df}{dx} = (f(x))^2##, so maybe the derivation arose in the context of the inverse square law of gravitation. That's really the only thing I can think of that's remotely related to what OP is saying.

Edit: maybe something like ##\frac{d^2u}{dr^2} \propto \frac{k}{r^2}##, therefore ##\frac{du}{dr} \propto \frac{k}{r}##? I dunno.
I started digging through some of my notes a few hours ago to see if I could find exactly what he did, but no success in finding it so far. It wasn't the law of gravitation derivation, I think it had to do with energy of objects in a bound orbit? And actually what he did was just take the square root of everything on the right hand of the equals sign; the original function wasn't squared, we just treated it as though to was, so everything ended up being to the one-half power. (I apologize if my over simplification of the derivation in the original post is muddying the question too much.) Taking the root was his shortcut for avoiding integration.

I'll keep digging through notes when I have access to them and see if I can track the derivation down in its entirety. It's still bothering me because what he did worked in the end, but I don't understand why.

ryaamaak said:
In my classical mechanics course, the professor did a bit of algebraic wizardry in a derivation for one of Kepler's Laws where a second derivative was simplified to a first derivative by taking the square root of both sides of the relation. It basically went something like this:
$$\frac{d^2 u}{dx^2} = \big( f(x)\big)^2\\ \frac{d u}{dx} \frac{d u}{dx} = \big( f(x)\big)^2 \\\frac{d u}{dx} = f(x)$$.
Why could my professor do this? How does this make sense in a physical sense? Or in a mathematical sense? What are the cases for where this trick won't work? Any clarification is much appreciated.
The problem with this derivation is first and foremost this:

On the left, there is a differential operator. On the right, there is a function.
Now, multiplication by a function is in general also a linear operation, but there is no function that is the differential operator in general. Why? It is pretty simple:
Assume there is a function g that for all functions f does f' = g*f.
$$u(x)=e^x\\ \frac{d u}{dx} = e^x\\ \frac{d }{dx} u(x) = 1 u(x)\\ g=1$$.
$$u(x)=e^{2x}\\ \frac{d u}{dx} = 2e^{2x}\\ \frac{d }{dx} u(x) = 2 u(x)\\ g=2$$.

sophiecentaur

## 1. What is a "undo" second derivative with square root?

An "undo" second derivative with square root refers to the process of finding the original function from its second derivative, where the function contains a square root term.

## 2. How do you calculate an "undo" second derivative with square root?

To calculate an "undo" second derivative with square root, you first need to take the derivative of the original function twice. Then, you can solve for the original function by isolating the square root term and applying the inverse operation.

## 3. Why is it important to find the "undo" second derivative with square root?

Finding the "undo" second derivative with square root allows us to understand the nature of the original function and make predictions about its behavior. It also helps in solving real-world problems related to rates of change and optimization.

## 4. What are some common mistakes when calculating an "undo" second derivative with square root?

Some common mistakes when calculating an "undo" second derivative with square root include forgetting to take the derivative of the original function twice, not isolating the square root term properly, and making errors in algebraic manipulations.

## 5. Can the "undo" second derivative with square root be applied to any function?

No, the "undo" second derivative with square root can only be applied to functions that contain a square root term. It cannot be used for functions that do not involve a square root term, as the process of taking the inverse operation would not be valid.

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