Derivative problem -- Chain rule

[LagrangeEuler]

Homework Statement

Derivative question
f=f(x) and x=x(t)
then in one book I find
[tex]\frac{d}{dx}\frac{df}{dt}=\frac{d}{dx}(\frac{df}{dx}\frac{dx}{dt})[/tex]
[tex]=\frac{dx}{dt} \frac{d^2 f}{dx^2} [/tex]

Homework Equations

The Attempt at a Solution

Not sure why this is correct? [itex]\frac{dx}{dt}[/itex] can depend of [itex]f[/itex] for example. Right?

 

[Fredrik]

The chain rule is ##\frac{df}{dt}=\frac{df}{dx}\frac{dx}{dt}##. I prefer the notation ##(f\circ x)'(t)=f'(x(t))x'(t)##. The notation ##\frac{d}{dx}\frac{df}{dt}## doesn't make sense to me. ##\frac{df}{dt}## should mean ##(f\circ x)'(t)##. Since there's no function that takes x to ##(f\circ x)'(t)##, it makes no sense to have ##\frac{d}{dx}## act on ##\frac{df}{dt}##.

Where did you see that claim? What book and page number? If possible, your best option is to link directly to the page at google books.

 

[LagrangeEuler]

In physics for example you often have trajectory and you want to find velocity. For example [tex]y(x)=x^2[/tex] and you want to find [tex]\upsilon_y[/tex]. You know that [tex]\upsilon_x=t[/tex]
[tex]\upsilon_y=\frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}=[/tex]
[tex]=\upsilon_x \frac{dy}{dx}=2xt[/tex].
So calculations are obtain by multiplication with [tex]\frac{dx}{dx}[/tex].

 

[Fredrik]

I guess your point must be that ##\frac{d}{dx}\frac{dy}{dt}## should make sense in your example, since ##\frac{dy}{dt}=2xt##. This equation is really saying that ##(y\circ x)'(t)=2x(t)t##. I wouldn't say that it makes sense to apply ##\frac{d}{dx}## to that right-hand side.

I need to think a little bit more. Unfortunately I have to leave the computer for about an hour right now. It would still help if you could post that book refereence.

 

[Ray Vickson]

In physics for example you often have trajectory and you want to find velocity. For example [tex]y(x)=x^2[/tex] and you want to find [tex]\upsilon_y[/tex]. You know that [tex]\upsilon_x=t[/tex]
[tex]\upsilon_y=\frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}=[/tex]
[tex]=\upsilon_x \frac{dy}{dx}=2xt[/tex].
So calculations are obtain by multiplication with [tex]\frac{dx}{dx}[/tex].

No. You did not use ##dx/dx## anywhere in the above calculation, but you did use ##dy/dx## and ##dx/dt##.

 

[LagrangeEuler]

[tex]\upsilon_y=\frac{dy}{dt}=\frac{dy}{dt}\frac{dx}{dx}=\frac{dy}{dx}\frac{dx}{dt}[/tex]

 

[Fredrik]

[tex]\upsilon_y=\frac{dy}{dt}=\frac{dy}{dt}\frac{dx}{dx}=\frac{dy}{dx}\frac{dx}{dt}[/tex]

That thing in the middle is not an intermediate step that you can use to prove that the thing on the left is equal to the thing on the right. There is a way to make sense of dx,dy,dt, but if you use it, the two dx's on the right-hand side aren't the same, and the dy on the left isn't the same as the dy on the right.

 

[nylonsmile]

Try applying the product rule to: $$\frac{d}{dx}\left( \frac{df}{dx}\frac{dx}{dt}\right) $$

 

[Fredrik]

Try applying the product rule to: $$\frac{d}{dx}\left( \frac{df}{dx}\frac{dx}{dt}\right) $$

The notation ##\frac{d}{dx}(\text{something})## means "the value at x of the derivative of the function that takes x to (something)". But "the function that takes x to ##(f\circ x)'(t)x'(t)##" is ill defined.

Edit: I should have said ##f'(x(t))x'(t)## or ##(f\circ x)'(t)##, not ##(f\circ x)'(t)x'(t)##.

 

[nylonsmile]

The notation ##\frac{d}{dx}(\text{something})## means "the value at x of the derivative of the function that takes x to (something)". But "the function that takes x to ##(f\circ x)'(t)x'(t)##" is ill defined.

Hmm yeah... But I feel like you could use a difference quotient to find ##\frac{d}{dx}\left( \frac{df}{dx}\frac{dx}{dt}\right) ## but I could be wrong. If I had the functions ##x(t)## and ##f(x)## it seems like I could make a computer work out the required function. Do something like:

1. At ##x##, find ##\frac{f(x+\delta x) - f(x)}{\delta x}##
   
2. At ##x##, find ##\frac{x(t+\delta t) - x(t)}{\delta t}##
3. Multiply these two numbers together.

Is that not the required function? Is the problem that we don't know what time to use?

I am confused.

What is wrong with doing this?: $$\frac{d}{dx}\left( \frac{df}{dx}\frac{dx}{dt}\right) = \frac{dx}{dt}\frac{d^2f}{dx^2} + \frac{df}{dx}\frac{d^2x}{dxdt}$$

 

[Fredrik]

What is wrong with doing this?: $$\frac{d}{dx}\left( \frac{df}{dx}\frac{dx}{dt}\right) = \frac{dx}{dt}\frac{d^2f}{dx^2} + \frac{df}{dx}\frac{d^2x}{dxdt}$$

The left-hand side doesn't make sense.

Hmm yeah... But I feel like you could use a difference quotient to find ##\frac{d}{dx}\left( \frac{df}{dx}\frac{dx}{dt}\right) ## but I could be wrong. If I had the functions ##x(t)## and ##f(x)## it seems like I could make a computer work out the required function. Do something like:

1. At ##x##, find ##\frac{f(x+\delta x) - f(x)}{\delta x}##
   
2. At ##x##, find ##\frac{x(t+\delta t) - x(t)}{\delta t}##
3. Multiply these two numbers together.

Is that not the required function? Is the problem that we don't know what time to use?

The problem is that the number obtained in 3 (if we take the appropriate limits in 1 and 2), is completely determined by the value of t, not by the value of x. The only x that appears in that expression is a function, not a number.

 

[nylonsmile]

The left-hand side doesn't make sense.The problem is that the number obtained in 3 (if we take the appropriate limits in 1 and 2), is completely determined by the value of t, not by the value of x. The only x that appears in that expression is a function, not a number.

Yeah I understand the problem now. Could we just imagine the whole thing is a function of t?

 

[Fredrik]

Yeah I understand the problem now. Could we just imagine the whole thing is a function of t?

Yes, ##(f\circ x)'## is a function, and ##(f\circ x)'(t)## is a function of t (i.e. its value is determined by the value of t).