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Finding the Derivative

  1. Oct 20, 2011 #1
    1. The problem statement, all variables and given/known data
    If h(2)=4 and h'(2)=-3, find
    [tex]\left.\frac{d}{dx}\frac{h(x)}{x}\right|_{x=2}[/tex]


    2. Relevant equations
    n^n-1 (power rule)

    3. The attempt at a solution
    I don't know how to get this started. It seems like I am having trouble with derivatives. I can do simple derivatives with the power rule, product rule, and quotient rule, but I do not know what the line on the right means, nor do I understand what the d/dx times the quantity of h(x)/x means.
     
    Last edited by a moderator: Oct 20, 2011
  2. jcsd
  3. Oct 20, 2011 #2
    It means find the derivative of f(x)/x when (or, at the point where) x equals 2.
     
  4. Oct 20, 2011 #3

    SammyS

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    d/dx times the quantity of h(x)/x means: the derivative of h(x)/x

    the line on the right means: evaluate the derivative at x = 2
     
  5. Oct 20, 2011 #4
    SO the derivative would be -3/4?
     
  6. Oct 20, 2011 #5

    SammyS

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    No. Are you using the quotient rule?

    Helps to show your work.
     
  7. Oct 20, 2011 #6
    Um, h(x)=4, because x=2 so h(2)=4; x=2, so I get 1/2. But you don't understand how you would the the quotient rule using that value. There is the product rule if you rearrange the formula to 1(2)^-1
    I'm sorry, I'm a little confused.
     
  8. Oct 20, 2011 #7

    SammyS

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    (Don't put 2 in for x just yet.)

    What is the derivative of [itex]\displaystyle \frac{h(x)}{x}[/itex] , using the quotient rule?

    If that doesn't make sense, then what is the derivative of [itex]\displaystyle \frac{h(x)}{g(x)}\,?[/itex]
     
  9. Oct 20, 2011 #8

    Mark44

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    No, you can't say this. You don't know what h(x) is, only what its value is at a particular x value.
    Forget the numbers for now.

    1. Find the derivative of h(x)/x. I would use the quotient rule.
    2. Evaluate the derivative you found in #1 at x = 2.
     
  10. Oct 20, 2011 #9
    [itex]\frac{x[h'(x)]-h(x)x'}{x^{2}}[/itex]
     
  11. Oct 20, 2011 #10

    Mark44

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    What's x' ?
    Simplify what you have.

    Then evaluate everything at x = 2.
     
  12. Oct 20, 2011 #11
    x=2
    [2(-3)-4(0)]/2^2 =-3/2
     
  13. Oct 20, 2011 #12

    Mark44

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    Not quite, but you're close. What's x'? (It's not 0.)
     
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