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Quick derivative

  1. Feb 7, 2005 #1
    i am having trouble with a fairly easy derivative - and was wondering if someone could show me the steps how to find this?

    Find this derivative algebraically

    f(x) = x^3 (x cubed) at x= -2

    The answer in the back of the book says the derivative is 12 - but I did the work and got 4. Please help!
  2. jcsd
  3. Feb 7, 2005 #2


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    Thy shalt not forget to multiply by the exponent... i.e.

    [tex](x^a)' = ax^{a-1}[/tex]
  4. Feb 7, 2005 #3
    Consider the function [tex] x^4 [/tex] Then [tex] \frac{dy}{dx} = 4x^{3} [/tex]

    Do the same for your function
  5. Feb 7, 2005 #4


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    Use the definition:Denote the derivative in the point "-2" by D.Then:

    [tex] D=:\lim_{x\rightarrow -2}\frac{f(x)-f(-2)}{x-(-2)}=\lim_{x\rightarrow -2}\frac{x^{3}+8}{x+2}=\lim_{x\rightarrow -2} x^{2}-2x+4 =+12 [/tex]

    ,where i made use of the identity:

    [tex] a^{3}+b^{3}=(a+b)(a^{2}-ab+b^{2}) [/tex]

  6. Feb 7, 2005 #5
    dextercioby did it through definition of a derivative. if you are lazy like me, you can do it a shorter way!
    if you have a polynomial function, and you want to find the derivative of it use the fact that if [itex]y=x^n[/itex], then [itex]\frac{dy}{dx} = nx^{n-1}[/itex].
    now, evaluate [itex]3x^2[/itex] at -2.
    that's the lazy way. :smile:
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