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Chain Rule

  1. Feb 9, 2007 #1
    When do you know to use the chain rule??

    I really don't understand what this means: dr/dt = dr/ds * ds/dt

    Like in one example in the book they have:
    r(s) = cos2si + sin2sj + e^-3sk where s = t^4

    Then they find dr/dt using the chain rule. My question is why can't you just differentiate it normally by the individual components? How would that be different?
  2. jcsd
  3. Feb 10, 2007 #2
    You use the chain rule when one function is the argument of the other:

    For example: lets say we have the function f(x) = (sin(x) + x^2)^3

    We differentiate this by asking our selves a simpler question. How would we differentiate x^3. Using the power rule we would get 3x^2. Sadly though f(x) isn’t x^3 it’s (sin(x) + x^2)^3. What the chain rule tells us is that we can differentiate this the same we would differentiate x^3 (pretending (sin(x)+x^2) is x), we just need to tack the derivative of the inside stuff on the end.

    So the derivative of sin(x) + x^2 is cos(x) +2x

    So we get the derivate of (sin(x) + x^2)^3 as [3(sin(x) + x^2)^2]*( cos(x) +2x)
  4. Feb 10, 2007 #3

    Gib Z

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    What JonF said is pretty much correct. Heres a way to understand why

    [tex]\frac{dy}{dx}= \frac{dy}{du} \cdot \frac{du}{dx}[/tex].

    If we want the rate of change of something, and it actually involves 2 functions, we use the chain rule. Sometimes it is also called the function of a function rule.

    There are 2 functions, each with their own rates of change. One is the inner function, eg (2si). That is the function f(s)=2si. Then the outer function is cos(2si), or cos(f(s)). If we want the TOTAL rate of change, we take the product of the two rates of change. so we let 2si=u. Then the function can be said to be cos u. dy/du= -sin u. du/dx= 2i. Replace the u with 2si again and your done.

    We can see the rule works this way: I'm walking at a certain speed. Someone else is driving 2 times as fast as me. A plane is going 2 times as fast as the car. How do we find the total rate of change, basically how fast the plane is compared to the walker, we multiply the rates of change!! This is the chain rule in action!!!

    Have fun :p
  5. Feb 10, 2007 #4


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    What do you mean by "differentiate it normally by the individual components"? Using the chain rule has nothing to do with the components. They suggest using the chain rule because of that "2s" in the trig functions and the "-3s" in the exponential.

    How would YOU differentiate cos 2s "normally"? If you would think "the derivative of cosine is negative sine and then I multiply the the derivative of 2s which is 2 to get -2cos 2x, then you Are using the chain rule on each individual component.
  6. Feb 10, 2007 #5

    matt grime

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    Ok, how do you differentiate cos(t^4)? You don't know how to do this without the chain rule. How do you know to use the chain rule? Because it is the composition of functions that you do know how to differentiate: f(x)=x^4 and g(x)=cos(x).
  7. Feb 10, 2007 #6

    D H

    Staff: Mentor

    Sure we do. Use the Weierstrass definition of the derivative as a limit (Cringe):

    [tex]\frac d{dt}\left(\cos(t^4)\right) =
    \lim_{\Delta t\to 0} \frac{\cos((t+\Delta t)^4)-\cos(t^4)}{\Delta t}[/tex]

    After way too much work (trig identities, binomial expansion, small angle approximation), one arrives at

    [tex]\frac d{dt}\left(cos(t^4)\right) = -4t^3 \sin(t^4)[/tex]

    Of course, one could just use the chain rule and get the same answer quickly. We use the chain rule because it is a lot easier than using the formal, limit-based definition of the derivative.
  8. Feb 10, 2007 #7

    matt grime

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    Was the 'you' in my sentence confusing you, D H?
  9. Feb 10, 2007 #8

    D H

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    No. I was simply expounding on why the chain rule is useful. Who would ever want to revert to limit-based calculation of a derivative when a few short-hand techniques like the chain rule, product rule, quotient rule, etc. make the job of computing derivatives easy?
  10. Feb 10, 2007 #9

    Gib Z

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    I never knew that Weierstrass made that definition, I've always been told that the expression was known as Newton's Quotient Expression or something like that, and that Newton made that one.
  11. Feb 10, 2007 #10

    Gib Z

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    BTW everyone seems to be confused when the OP said differently normally, Im pretty sure he meant that if he saw cos(BLAH) his derivative is -sin(BLAH), no matter what his BLAH is.
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