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Difference Quotient Isn't working.

  1. Jul 27, 2010 #1
    1. The problem statement, all variables and given/known data
    Find the derivative of f using the differance quotient and use the derivative of f to determine any points on the graph of f where the tangent line is horizontal.


    2. Relevant equations

    3. The attempt at a solution

    [tex]\lim_{\Delta x\rightarrow0}=\frac{3(x+\Delta x)^3-9(x+\Delta x)-3x^3+9x}{\Delta x}[/tex]

    [tex]\lim_{\Delta x\rightarrow0}=\frac{3(x^3+3x^2\Delta x+3x\Delta x^2+\Delta x^3)-9x-9\Delta x-3x^3+9x}{\Delta x}[/tex]

    [tex]\lim_{\Delta x\rightarrow0}=\frac{3x^3+9x^2\Delta x+9x\Delta x^2+3\Delta x^3-9x-9\Delta x-3x^3+9x}{\Delta x}[/tex]

    The '3x3's and '9x's cancel out and you are left with

    [tex]\lim_{\Delta x\rightarrow0}=\frac{9x^2\Delta x+9x\Delta x^2+3\Delta x^3-9\Delta x}{\Delta x}[/tex]

    Then if you take the limit as delta x approaches 0 you will get

    [tex]9x^2+9x-6[/tex] I know the answer is [tex]9x^2-9[/tex] via the rules of differentiation.

    What am I doing wrong with this example?
  2. jcsd
  3. Jul 27, 2010 #2

    Gib Z

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    Homework Helper

    Check the bolded step again.
  4. Jul 27, 2010 #3
    You can cancel factors of [itex]\Delta x[/itex] from each term in the numerator with the one in the denominator. What are you left with after this cancellation? What is the limit of this expression when [itex]\Delta x \rightarrow 0[/itex]?
  5. Jul 27, 2010 #4
    [tex]\lim_{\Delta x\rightarrow0}=\frac{9x^2\Delta x+9x\Delta x^2+3\Delta x^3-9\Delta x}{\Delta x}[/tex]

    Ok, so you would get this after taking the limit.

    [tex]=9x^2+9x\Delta x+3\Delta x^2-9[/tex]

    what happens to the delta x terms though? Is delta x just such a small number [tex]\epsilon[/tex] that you can say they are practically 0 and anything multiplied by 0 is 0?
  6. Jul 27, 2010 #5
    No. Please review limits before you go on to derivatives.
  7. Jul 27, 2010 #6
    Nevermind you would simply factor out a delta x out of the numerator then you have 2 'delta x's that cancel out. Then you take the limit and you get 9x2-9. Otherwise its in indiscriminate form.
  8. Jul 28, 2010 #7


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    Science Advisor

    I think you mean "indeterminant" form.
  9. Jul 28, 2010 #8
    I think you meant "indeterminate" form.
  10. Jul 28, 2010 #9


    Staff: Mentor

    HallsOfIvy was being indiscriminate.:smile:
  11. Jul 28, 2010 #10


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    Homework Helper

    Hahaha gold :rofl:
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