# Solving Difficult Derivatives: Need Help!

• smellymoron
In summary, the conversation includes a discussion about working with derivatives, specifically the forms (1) d/dx (x')^2 and (2) d/dx' (x). The participants also discuss the general solutions to these problems and how they relate to Euler-Lagrange problems. The conversation also delves into the concept of directional functional derivatives and the application of the Euler trick to find extrema. Finally, there is a brief discussion about the value of \frac{dx'}{dx} and how it relates to the second derivative of x with respect to t and the first derivative of x with respect to t.
smellymoron
Hi there
I'm working with some derivatives which I am having a lot of trouble with.
Here are the 2 forms I'm stuck on:

(1) d/dx (x')^2 , where x is a function of t and x' is the first derivative of x with respect to t.

(2) d/dx' (x) , same conditions as above, never seen one of these before.

Can anyone help me with the general solutions to these problems?
I think the answer to the first one is 2x'' where x'' is the second derivative of x with respect to t, but I'm really not sure and I have no idea with the second one.

Thanks for any help provided.

Last edited:
What is

$$\frac{dx'}{dx}$$

equal to...?

Daniel.

dex, say

$$x'(t) = t^2$$

Then

$$\frac{dx'(t)}{dt} = 2t$$

but

$$\frac{dx'(t)}{dx} = 0$$

?

That, unfortunately, is half the problem.
x(t) is undefined so I don't even have an example to work with.
I was hoping someone would have come across problems like this before, and would know the general form of how to solve them.
If it helps at all this is to solve an Euler-Lagrange problem.

Great,then u've been already told that

$$\frac{\partial q^{i}}{\partial \dot{q}^{j}}=0$$

and

$$\frac{\partial q^{i}}{\partial q^{j}}=\delta^{i}_{j}$$

Daniel.

Thanks dexter.
I hadn't seen either of those before. All I've got is some illegible notes to work from. I understand the first one now but what exactly is that on the right hand side in the second one? And what do the i's and j's represent?

Upper indices.They label degrees of freedom.That is delta-Kronecker.I'm sure u're familiar with it.

Daniel.

Hm, we are going here in heavy math ?

I think there are 2 ways of giving an answer at least :

a) week-end mathematician : you derive with respect to a function, but I don't care, since I do it a la physicist as computing differentials :

1) $$\frac{d (x'(t)^2)}{dx(t)}=\frac{2x'(t)x''(t)dt}{x'(t)dt}=2x''(t)$$
2) $$\frac{dx(t)}{dx'(t)}=\frac{x'(t)dt}{x''(t)dt}=\frac{x'(t)}{x''(t)}$$

b) Heavy math jam : you have to use the "directional" functional derivative (Gateaux), and your notation is not correct with these symbols : $$d/dx\rightarrow D_{x}[F]$$

Since x'(t)^2 is a functional of x as well as x(t) is a functional of x'.

NB: direction and place have no sense here since we speak about functions.

Reminder : the definition of the functional derivative in the "direction" v of a functional F at the "place" f:

$$D_{v}F[f]=\lim_{\epsilon\rightarrow 0}\frac{F[f+\epsilon v]-F[f]}{\epsilon}$$

Let's compute that way :

first :

1) F[x]=x'(t)^2
2) G[x]=x(t)

then :

1) $$\lim_{h->0}{\frac{(x(t)+hv(t))'^2-x'(t)^2}{h}=\lim_{h->0}\frac{(x'(t)+hv'(t))^2-x'(t)^2}{h}$$
$$=\lim_{h->0}\frac{2hx'(t)+h^2v'(t)^2}{h}$$
$$=2x'(<a style='text-decoration: none; border-bottom: 3px double;' href="http://www.serverlogic3.com/lm/rtl3.asp?si=22&k=t%20v" onmouseover="window.status='<a style='text-decoration: none; border-bottom: 3px double;' href="http://www.serverlogic3.com/lm/rtl3.asp?si=22&k=t%20v" onmouseover="window.status='<a style='text-decoration: none; border-bottom: 3px double;' href="http://www.serverlogic3.com/lm/rtl3.asp?si=22&k=t%20v" onmouseover="window.status='t)v'; return true;" onmouseout="window.status=''; return true;">t)v</a>'; return true;" onmouseout="window.status=''; return true;">t)v</a>'; return true;" onmouseout="window.status=''; return true;">t)v</a>'(t)$$

Now replace the arbitrary function v(t) by x(t) : you get :

$$D_xF[x]=2x'(t)^2$$

2) $$D_vG[x]=\lim_{h->0}\frac{x(t)+hv(t)-x(t)}{h}=v(t)$$

Replace v(t)=x'(t) you get D_x'G[x]=x'(t)

So the answer depends on how you see the problem :

a) you see the all the function as function of the parameter t, and apply physicist differentiation

b) you see the derivative of a functional towards a function and apply Gateaux derivative

REM : I have problems with the tex output

Last edited:
In Lagrange formalism,the generalized coordinates & velocities are independent variables,therefore differentiaition of coordinates wrt velocities yields 0...

Daniel.

I don't think the first question was related to mechanics, but was just a mathematical aspect.

However, in Lagrangian mechanics, the variables q, q' are not independent...I suppose you meant Hamiltonian mechanics with (q,p) ??

To smellymoron : the right answer is given by the Gateaux derivatives (every other calculation or formalism is a mathematical nonsense) : so that the answers should be :

1) 2x'(t)^2
2) x'(t)

Take a look at every book on variational calculus, you will see they use the Euler trick (Gateaux derivative) to find the extrema of the action with respect to the motion :

in this case you have the action : $$S[x]=\int_a^b L(x,x',t)dt$$

where L is the Lagrangian which is mathematically a functional of the motion and it's derivatives. The Euler-Lagrange equations are then obtained by finding the extrema of the action : suppose X is an extremum, then for every function n, such that n(a)=n(b)=0 we have :

$$\lim_{h->0}\frac{S[X+hn]-S[X]}{h}=0$$
$$=\lim_{h->0}\frac{1}{h}\int_a^b\frac{\partial L}{\partial x}hn(t)+\frac{\partial L}{\partial x'}hn'(t)dt$$ the higher order terms in h are irrelevant. Then by integrating by parts :

$$\int_a^b\frac{\partial L}{\partial x}-\frac{d}{dt}\left(\frac{\partial L}{\partial x'}\right)n(t)dt=0 \forall n(t)$$

This gives you the Euler-Lagrange equation since the integrand has to vanish. To get the Hamiltonian version you just apply the Legendre transformation.

whozum said:
dex, say

$$x'(t) = t^2$$

Then

$$\frac{dx'(t)}{dt} = 2t$$

but

$$\frac{dx'(t)}{dx} = 0$$

?

Well, why should this be 0 ?? I can just do the following :

$$x(t)=\int x'(t)dt=\frac{t^3}{3}+C$$

Now I do the change of variable $$t=t(x)=\sqrt[3]{3(x-C)}$$
which implies :

$$x'(x)=x'(t(x))=t(x)^2=(3(x-C))^{2/3}$$

so that in this special case :

$$\frac{dx'}{dx}=2\frac{1}{\sqrt[3]{3(x-C)}}$$

BTW: You see that this is exactly x''(t)/x'(t) with t=t(x) ! Which is the week-end mathematician method presented before.

Last edited:
They are independent in Lagrangian formalism.Read more into it.

Daniel.

## 1. What are derivatives and why are they important?

Derivatives are mathematical tools used to find the rate of change of a function at a specific point. They are important because they have many applications in fields such as physics, engineering, economics, and statistics.

## 2. How do I approach solving difficult derivatives?

First, make sure you have a good understanding of basic derivative rules and techniques. Then, carefully analyze the given function and apply the appropriate rules and techniques. It's also helpful to break down the problem into smaller steps and simplify as much as possible.

## 3. What are some common mistakes to avoid when solving derivatives?

One common mistake is forgetting to use the chain rule when dealing with composite functions. Another is making algebraic errors, such as forgetting to distribute or simplifying incorrectly. It's also important to double-check your final answer and make sure it matches the given function.

## 4. How can I improve my skills in solving derivatives?

Practice, practice, practice! The more problems you solve, the more comfortable you will become with different types of derivatives. It's also helpful to seek out additional resources, such as textbooks, online tutorials, and practice problems.

## 5. Are there any tips or tricks for solving difficult derivatives?

One strategy is to look for patterns in the given function and see if it can be rewritten in a simpler form. Another tip is to use visual aids, such as graphs or diagrams, to better understand the behavior of the function. It's also helpful to work through problems step by step and not rush through the process.

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