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Math Help - Chain Rule - Proof Using...

  1. #1
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    Chain Rule - Proof Using...

    I have the question, and part of my work... I can't figure out where to go from here. Bear in mind, this is university level calculus, grade 12... We don't do this type of question too often, so I might need the extra step included in between your regular steps.

    Givens:
    y=f(x)
    f(a)=a
    f '(a)=b
    g= f o f o f o f o f o f o f o f o f (if I'm using the wrong symbol here, I mean to say f at f at f at f at ....... and so on, 10 times)

    Question: find g '(a)

    Work/Answer:
    g(x)=(f(x))^10
    g(a)=(f(a))^10
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  2. #2
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    Quote Originally Posted by mike_302 View Post
    I have the question, and part of my work... I can't figure out where to go from here. Bear in mind, this is university level calculus, grade 12... We don't do this type of question too often, so I might need the extra step included in between your regular steps.

    Givens:
    y=f(x)
    f(a)=a
    f '(a)=b
    g= f o f o f o f o f o f o f o f o f (if I'm using the wrong symbol here, I mean to say f at f at f at f at ....... and so on, 10 times)

    Question: find g '(a)

    Work/Answer:
    g(x)=(f(x))^10
    g(a)=(f(a))^10
    ?? The problem asked you to find g'(a). Why did you give g(a)?

    Try a simple case first. Suppose g(x)= f(f(x)). Then g'(x)= f'(f(x))f'(x). Then g'(a)= f'(f(a))f'(a)= f'(a)f'(a)= b^2.

    If g(x)= f(f(f(x))), then g'= f'(f(f(x))(f(f(x))' so g'(a)= f'(f(f(a))(b^2)= f'(a)(b^2)= b^3.

    Now, make a wild guess at what your answer should be!
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  3. #3
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    Quote Originally Posted by mike_302 View Post
    I have the question, and part of my work... I can't figure out where to go from here. Bear in mind, this is university level calculus, grade 12... We don't do this type of question too often, so I might need the extra step included in between your regular steps.

    Givens:
    y=f(x)
    f(a)=a
    f '(a)=b
    g= f o f o f o f o f o f o f o f o f (if I'm using the wrong symbol here, I mean to say f at f at f at f at ....... and so on, 10 times)

    Question: find g '(a)

    Work/Answer:
    g(x)=(f(x))^10
    g(a)=(f(a))^10
    Define f_n(x) = \underbrace{f \circ  f \circ \cdots \circ f}_{n}(x).

    Observe a couple of facts:
    1. f_{n}(x) = f \circ f_{n-1}(x) for n > 1, and
    2. f_n(a) = a for all n.

    So g(x) = f_n(x).

    What is f_n'(x)?
    Well it is \left(f \circ f_{n-1}\right)'(x). By the chain rule, this is f' \circ f_{n-1}(x) \cdot f_{n-1}'(x).

    We continue this process until the subscript is 1 in the rightmost term.

    f_n'(x) = f' \circ f_{n-1}(x) \cdot f' \circ f_{n-2}(x) \cdots f' \circ f_{2}(x) \cdot f' \circ f_1(x) \cdot f_1'(x).

    By observation 2 above, then:
    f_n'(a) = \underbrace{f'(a) \cdot f'(a) \cdots f'(a) \cdot f'(a)}_{n}.

    Can you take it from there?
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  4. #4
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    Ahhh! Clever. I did not think to start off with only the pair and go from there. I was trying to do it with all ten in there.

    Okay, that's an interesting way of tackling it.

    Appreciated!
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