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AP calculus AB, 2 related problems

  1. Oct 4, 2011 #1
    1. The problem statement, all variables and given/known data
    First problem:

    Data in the form of a table is given for all integers x in [-2,3], f(x), g(x), f'(x), and g'(x)
    Given that h(x) = g(2/x), find h'(2/3)

    Second problem:

    Given g(x) = ∫2-3xx2 f(t) dt, and h(x) = g(x2), find the derivative of h'(x)? Lower bound on the integral is 2-3x and upper bound is x2.


    2. Relevant equations



    3. The attempt at a solution

    First problem:

    Do we or do we not apply the chain rule? Since h(x) = g(2/x), then does h'(x) = g'(2/x)?

    Second problem:

    We can simplify g(x) to the integral f(t) evaluated at x2 - the integral of f(t) evaluated at 2-3x. If g(x) is equal to this, then the derivative of it is 2x*f(x2) - 3*f(2-3x), since where taking derivative wrt x. Then finally, we plug in x2 wherever there is an x to get h'(x) (=g'(x2)
     
  2. jcsd
  3. Oct 4, 2011 #2

    SammyS

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    First Problem: Yes, you apply the chain rule.

    The chain rule gives [itex]\displaystyle h'(x)=g'\left(\frac{2}{x}\right)\cdot\frac{d}{dx} \left(\frac{2}{x}\right)\,.[/itex]

    Second problem:

    Suppose [itex]F(x)=\int\ f(x)\ dx[/itex], then [itex]F\,'(x) = f(x)\,.[/itex]

    Your integral is then given by: [itex]\displaystyle \int_{x^2}^{2-3x}\ f(t)\ dt = F(x^2)-F(2-3x)[/itex]

    Take the derivative of that. Don't forget the chain rule & that [itex]F\,'(x) = f(x)\,.[/itex]
     
  4. Oct 4, 2011 #3
    Thanks for helping me to understand the first problem.

    For the second problem, I am wondering do we plug in x2 for x before or after we differentiate, since h(x) = g(x2)? What is the reason for this?
     
  5. Oct 4, 2011 #4

    Never mind, I figured it out. Thanks anyway for all the help!
     
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