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Given the following values of x, g(x) and g'(x), what is h'(2/3) if h(x)=g(2/x)?

  1. Oct 7, 2011 #1
    1. The problem statement, all variables and given/known data
    x g(x) g'(x)
    -2 -2 2
    -1 0 1
    0 1 2
    1 3 4
    2 7 3
    3 9 2


    2. Relevant equations

    chain rule

    3. The attempt at a solution

    differentiate (2/x) and multiply that times g'(2/x). Plug in 2/3 into -2/(x^2), and one obtains -9/2. g'(2/((2/3)))= 2. The two multiplied together yields -9. Is that correct?
     
  2. jcsd
  3. Oct 7, 2011 #2

    SammyS

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    Please, Don't double-post questions.

    Where's the problem statement? .... Oh! It's there in the title line. (It's a good idea to repeat the question in the main text.)

    Yes, d/dx (g(2/x)) evaluated at x = 2/3 is g'(2/(2/3))∙(-2/(2/3)2).
     
  4. Oct 7, 2011 #3
    Sorry about not posting the problem statement in the main text. And I double-posted because it was first in the wrong category. This is calc., not pre-calc.
     
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