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Homework Help: Chain Rule Trig Derivative Problem

  1. Nov 8, 2012 #1
    1. The problem statement, all variables and given/known data

    Find the derivative of y = sin(πx)2

    2. Relevant equations

    Chain Rule: y' = f'(u) * u'

    3. The attempt at a solution

    (See attached image)

    The answer according to the textbook is 2π2xcos(πx)2. What am I doing wrong here?

    Attached Files:

  2. jcsd
  3. Nov 8, 2012 #2


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    I really can't tell what you are doing. What are using for f and u when you use the chain rule you stated?
  4. Nov 8, 2012 #3


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    Write your function as

    y = sin((πx)2)

    or as

    y = sin(π2x2)

    That may help you apply the chain rule correctly.
  5. Nov 8, 2012 #4
    Ok I did it by counting u as (π*x)2. When I took the derivative of that and made sin(πx)2 into cos(πx)2 I got the correct answer. But I don't see why you don't make it 2cos(πx)(2π2x). That's how I got my initial answer. I only was able to get the correct answer by trying different things until it worked.
  6. Nov 8, 2012 #5


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    What did you use for f and u to get 2cos(πx)(2π2x)??
  7. Nov 8, 2012 #6
    Well I just said what I used u as. I used cos u as f
  8. Nov 8, 2012 #7
    Have you ever tried www.wolframalpha.com it can help with basic computations like these.

    But as you know.... this is the equivalent of sin^2(∏x) where ∏ is a constant. so you apply the chain rule using U substitution. where U= sin(∏x). so you've got a double chain rule case....

    where U^2 derivative is 2ududx so you wind up with 2 sin(∏x)*cos(∏x)*∏ then you just get
  9. Nov 8, 2012 #8
    A good way to do this is to break it down so its easier to apply the chain rule for derivation.

    I'll assume we are considering the function [itex]f(x)=\sin((\pi x)^2)[/itex].

    To break this down, define the function [itex]S(x)=x^2[/itex].

    Also, define the function [itex]g(x)=\pi x[/itex].

    (Note that these [itex]x[/itex] parameters in these function definitions are not some same variable [itex]x[/itex], but just a dummy variable/parameter.)

    Therefore, [itex]f(x)[/itex] can be rewritten as [itex](\sin \circ S \circ g)(x)=\sin(S(g(x )))[/itex].

    Now, it is much easier to apply the chain rule, which turns out to be a triple application of it, here:

    [itex]f^{\prime}(x)=\sin^{\prime}(S(g(x))) \cdot \frac{d}{dx} (S(g(x)))[/itex]
    [itex]\therefore f^{\prime}(x) = \cos(S(g(x))) \cdot S^{\prime}(g(x)) \cdot \frac{d}{dx} (g(x))[/itex]
    [itex]\therefore f^{\prime}(x) = \cos(S(g(x))) \cdot (2 \cdot g(x)) \cdot g^{\prime}(x) \cdot \frac{d}{dx} (x)[/itex]

    I guess it might appear a bit tedious to define even more functions than you already have, but it makes application of the chain rule even more clear. Of course, once you get better at this, you'll be able to do this extra work mentally (I still do it sometimes, though, ^_^)
  10. Nov 8, 2012 #9


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    No, you didn't. If you had used f(u)=cos(u) then you would have have had f'(u)=(-sin(u)) in your answer. You've got a perfectly good formula for the chain rule, you just aren't using it.
  11. Nov 8, 2012 #10
    Practice makes perfect, I can see. Well thanks for the help. I think I understand this a little better now.
  12. Nov 8, 2012 #11
    haha, yeah, definitely. Yeah, I remember finding the chain rule a little weird. Just sit on it for a bit, and don't worry if it doesn't make "perfect sense" immediately.
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