1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Derivative problem -- Chain rule

  1. Nov 4, 2014 #1
    1. The problem statement, all variables and given/known data
    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]


    2. Relevant equations


    3. 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?
     
    Last edited by a moderator: Nov 4, 2014
  2. jcsd
  3. Nov 4, 2014 #2

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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.
     
  4. Nov 4, 2014 #3
    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].
     
  5. Nov 4, 2014 #4

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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.
     
    Last edited: Nov 4, 2014
  6. Nov 4, 2014 #5

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    No. You did not use ##dx/dx## anywhere in the above calculation, but you did use ##dy/dx## and ##dx/dt##.
     
  7. Nov 5, 2014 #6
    [tex]\upsilon_y=\frac{dy}{dt}=\frac{dy}{dt}\frac{dx}{dx}=\frac{dy}{dx}\frac{dx}{dt}[/tex]
     
  8. Nov 5, 2014 #7

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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.
     
  9. Nov 5, 2014 #8
    Try applying the product rule to: $$\frac{d}{dx}\left( \frac{df}{dx}\frac{dx}{dt}\right) $$
     
  10. Nov 5, 2014 #9

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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)##.
     
    Last edited: Nov 5, 2014
  11. Nov 5, 2014 #10
    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?: $$\frac{d}{dx}\left( \frac{df}{dx}\frac{dx}{dt}\right) = \frac{dx}{dt}\frac{d^2f}{dx^2} + \frac{df}{dx}\frac{d^2x}{dxdt}$$
     
  12. Nov 5, 2014 #11

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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.
     
  13. Nov 5, 2014 #12
    Yeah I understand the problem now. Could we just imagine the whole thing is a function of t?
     
  14. Nov 5, 2014 #13

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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).
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Derivative problem -- Chain rule
Loading...