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Expressing defined integral as composition of differentiable functions

  1. Oct 25, 2016 #1
    1. The problem statement, all variables and given/known data
    Let ##f(t)=\int_{t}^{t^2} \frac{1}{s+\sin{s}}ds,t>1.##Express ##f## as a composition of two differentiable functions ##g:ℝ→ℝ^2## and ##h:ℝ^2→ℝ##. In addition, find the derivative of ##f## (using the composition).

    2. Relevant equations


    3. The attempt at a solution
    Honestly, I have no proper idea how to approach this problem. I know what is being asked, but how to find such functions ##g## and ##f##? Let g(t)=(t,sint) and h(x,y)=1/(x+y)? A nudge in the right direction will be appreciated.
     
    Last edited: Oct 25, 2016
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  3. Oct 25, 2016 #2

    PeroK

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    What about ##g(t) = (t, t^2)##?

    PS I think I can see what is intended, but if the aim is to find ##f'## I don't see why you need to consider functions of more than one variable.
     
    Last edited: Oct 25, 2016
  4. Oct 26, 2016 #3
    The problem actually consists of two parts (a and b) and the part b is to find the derivative of ##f## using the composition. ##g(t)=(t,t^2)##, well, okay, but how to find such ##h##? ##h(x,y)=\int_{x}^{y} \frac{1}{s+\sin{s}}ds##? And in part b in thinking maybe it's the reversal chain rule? It's a bit confusing.
     
    Last edited: Oct 26, 2016
  5. Oct 26, 2016 #4

    PeroK

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    Yes, ##f(t) = h(g(t))## with those definitions.

    For your information, I would have split ##f## as follows:

    ##f(t) = f_1(t^2) - f_1(t)##

    Where ##f_1(t) = \int_0^{t} k(s)ds##

    Those are all differentiable functions of a single variable.
     
  6. Oct 26, 2016 #5
    Are there other possibilities?
     
  7. Oct 26, 2016 #6

    PeroK

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    There may be, but I think that is the one that the question setter intended.
     
  8. Oct 26, 2016 #7
    By the chain rule ##\frac{\partial{f}}{\partial{t}}(t)=\frac{\partial{h(g(t))}}{\partial{t}}=\frac{\partial{h}}{\partial{g_1}}\frac{\partial{g_1}}{\partial{t}}+\frac{\partial{h}}{\partial{g_2}}\frac{\partial{g_2}}{\partial{t}}##

    How to apply it in this case?
     
  9. Oct 26, 2016 #8

    PeroK

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    First, ##f## is a function of ##t##, so you can't have a partial derivative.

    You need to write:

    ##f(t) = g(x(t), y(t)) = g(t, t^2)##

    And, apply the chain rule to that.
     
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