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Math Help - Definition of derivative

  1. #1
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    Definition of derivative

    Please I need the steps for the solution.

    by using definition of derivative find the f'(x)

    the question is

    f(x)= 2/x^3

    Thank you in advance.
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  2. #2
    Super Member Matt Westwood's Avatar
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    Quote Originally Posted by Abdul View Post
    Please I need the steps for the solution.

    by using definition of derivative find the f'(x)

    the question is

    f(x)= 2/x^3

    Thank you in advance.
    From first principles:

    f'(x) = \frac {\frac 2 {(x+h)^3} - \frac 2{x^3}} h

    Expand that out and simplify, then let h tend to zero.
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  3. #3
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    I could not get the final answer correct.

    see what I did

    f'(x)= \frac {2x^3 -2x^3 -2h^3} {(h)(x+h)^3(x^3)}

    Then I got

    f'(x)= \frac {-2h^3} {(h)(x+h)^3(x^3)}

    What to do next?
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  4. #4
    Junior Member guildmage's Avatar
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    First, recall that f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h}.

    Now, from the given f\left( x \right) = \frac{2}{{{x^3}}}, you'll get f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{2{x^3} - 2{{\left( {x + h} \right)}^3}}}{{h{x^3}{{\left( {x + h} \right)}^3}}}.

    Whan you expand \frac{{2{x^3} - 2{{\left( {x + h} \right)}^3}}}{{h{x^3}{{\left( {x + h} \right)}^3}}}, you'll get \frac{{2{x^3} - 2{x^3} - 6x{h^2} - 6{x^2}h - 2{h^3}}}{{{x^6}h + 3{x^4}{h^3} + 3{x^5}{h^2} + {x^3}{h^4}}}.

    We can cancel an h from each of the terms in the numerator and denominator and we know that 2x^3 - 2x^3 = 0 and thus we have:

    f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{ - 6xh - 6{x^2} - 2{h^2}}}{{{x^6} + 3{x^4}{h^2} + 3{x^5}h + {x^3}{h^3}}}

    Since h tends to zero, each of the terms with an h will be zero. You'll be left with f'\left( x \right) = \frac{{ - 6{x^2}}}{{{x^6}}} =  - 6{x^{ - 4}}.

    You made a mistake in expanding \frac{{2{x^3} - 2{{\left( {x + h} \right)}^3}}}{{h{x^3}{{\left( {x + h} \right)}^3}}}. This is not equal to \frac{{2{x^3} - 2{x^3} - 2{h^3}}}{{h{{\left( {x + h} \right)}^3}{x^3}}} because - 2{\left( {x + h} \right)^3} \ne  - 2{x^3} - 2{h^3}.

    I hope this helps you.
    Last edited by guildmage; July 15th 2010 at 01:45 AM.
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  5. #5
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    Thanks a lot for your help. now I got where I made a mistake.
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  6. #6
    Super Member Matt Westwood's Avatar
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    Quote Originally Posted by Abdul View Post
    Thanks a lot for your help. now I got where I made a mistake.
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