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Power Rule

  1. Oct 6, 2016 #1
    The problem statement, all variables and given/known data
    y=5-1/x

    The attempt at a solution
    y'=5-1/xln5(-1)(x-2/(x^2)
    y'=-5-1/xx-2ln5/x2

    But apparently my book says that the x^-2 should be -x^-2
    That's what I don't understand. Some help please?
     
  2. jcsd
  3. Oct 6, 2016 #2

    Mark44

    Staff: Mentor

    Your thread title, "Power Rule," doesn't apply to the function in this problem, which is in fact an exponential function.
    Power rule: ##\frac d {dx}(x^n) = nx^{n - 1}##
    Chain rule form of power rule: ##\frac d {dx}(u^n) = nu^{n - 1} \cdot \frac {du}{dx}##
    Exponential function rule with chain rule: ##\frac d {dx}(e^u) = e^u \cdot \frac {du}{dx}##
    You have a mistake above -- that x-2 factor shouldn't be there.
    You have the minus sign in -1. The book apparently shows it with -x-2. It's like -2 * 5 versus 2 * (-5) -- same result.
     
  4. Oct 6, 2016 #3
    No, you misunderstand. I typed what I meant, minus the title error.
    The -5 is supposed to be there, but IN ADDITION there is a negative in front of the x in the solution.
     
  5. Oct 6, 2016 #4
    Solution from the book:

    y'=5-1/x(ln5)[-1*(-x-2)]
     
  6. Oct 6, 2016 #5
    Wolfram verifies the result from the book.
     
  7. Oct 6, 2016 #6

    Ray Vickson

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    Homework Helper

    The book's solution is correct (although I would have written ##1/x^2## or ##x^{-2}## instead of ##(-1)(-x^{-2})##. Just use the fact that for constants ##c## and ##n## we have
    $$\frac{d}{dx} c x^n = c n x^{n-1}.$$
    Apply this to ##c=-\ln(5)## and ##n = -1##.
     
  8. Oct 6, 2016 #7
    I just don't understand where the second negative comes from. I understand the negative one as that's the sign of the exponent, but what's the negative doing on x^-2? (-x^-2)
    Because without the second negative sign to cancel out the first (the first being -1), my final answer is ending up as negative...
     
  9. Oct 7, 2016 #8

    Mark44

    Staff: Mentor

    ##y = 5^{-1/x} = (e^{\ln(5)})^{-1/x} = e^{-\ln(5)/x}##
    ##y' = e^{-\ln(5)/x} \cdot \frac d {dx}(-\ln(5)/x) = 5^{-1/x} \cdot \frac d {dx}(-\ln(5)/x)##

    What do you get for the part I didn't finish?
     
  10. Oct 7, 2016 #9
    I dunno I haven't got to derivatives of logs in my book yet :(
     
  11. Oct 7, 2016 #10

    Mark44

    Staff: Mentor

    -ln(5) is a constant that multiplies 1/x. Surely you've seen a rule on how to differentiate a constant times a function.
     
    Last edited: Oct 7, 2016
  12. Oct 7, 2016 #11
    Thanks for your reply. Could I please get a step by step method that doesn't use Euler's number
     
  13. Oct 7, 2016 #12

    Mark44

    Staff: Mentor

    ???
    Post #8 is nearly complete, and goes step by step. All you have to do is differentiate -ln(5)/x. The last part of the equation I showed in post #8 doesn't involve e.
     
  14. Oct 7, 2016 #13

    Mark44

    Staff: Mentor

  15. Oct 7, 2016 #14

    SammyS

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    No. We don't give step-by-step.

    ... but try it this way:

    Start with ##\ \displaystyle y=\left(\frac{1}{5} \right)^{1/x} \,.\ ##
     
  16. Oct 7, 2016 #15
    I've finished that problem I appreciate it. It seems I've been making dumb mistakes with the signs.
     
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