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Math Help - limit definition of the derivative...final exam review!

  1. #1
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    limit definition of the derivative...final exam review!

    i cant quite remember how to do these,

    Use the limit definition of derivative to calculate
    f'(x) where f (x) = 1/

    x^3



    i want to make sure i understand all the practice examples thoroughly for my exam. i have some previous questions i still need help with, so help on those or this is greatly appreciated!
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  2. #2
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    Quote Originally Posted by twostep08 View Post
    i cant quite remember how to do these,

    Use the limit definition of derivative to calculate
    f'(x) where f (x) = 1/

    x^3



    i want to make sure i understand all the practice examples thoroughly for my exam. i have some previous questions i still need help with, so help on those or this is greatly appreciated!
    I assume that you want to use the definition

    f'(x)=\lim_{h \to 0}\left(\dfrac{f(x+h)-f(x)}{h}\right)

    If so:

    f'(x)=\lim_{h \to 0}\left(\dfrac{\frac1{(x+h)^3}-\frac1{x^3}}{h}\right) = \lim_{h \to 0}\left(\dfrac{x^3-(x+h)^3}{hx^3(x+h)^3}\right)

    Expand the terms in the numerator and collect like terms:

    f'(x)= \lim_{h \to 0}\left(\dfrac{-3hx^2-3h^2x-h^3}{hx^3(x+h)^3}\right)

    Factor out h in the numerator:

    f'(x)= \lim_{h \to 0}\left(\dfrac{h(-3x^2-3hx-h^2)}{hx^3(x+h)^3}\right)

    Cancel h.

    Now h is approaching zero: Then the numerator becomes -3x^2 and the denominator becomes x^6

    Therefore

    f'(x)=\lim_{h \to 0}\left(\dfrac{\frac1{(x+h)^3}-\frac1{x^3}}{h}\right) = \dfrac{-3x^2}{x^6} = -\dfrac3{x^4}
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  3. #3
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    thats what i initially thought of doing.. is there anything else the question could mean becuase ive only heard that referred to as "the formal definition of the derivitave" i believe

    i certainly hope its the way you showed , because i understand that, but do you think its possible that the ? refers to another method?
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