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Math Help - Derivative using definition

  1. #1
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    Derivative using definition

    If f(x)=1/x^2 find f'(x) using the definition of the derivative.
    (note: If you instead use the power rule, f'(x)=nx^(n-1), you will get zero credit for this problem. It is useful check of your answer though.)

    The answer is -2/x^3 and the best I can get is -1/x^3.
    Could someone get me on the right track? If you can't help by tomorrow, don't bother because it is due tomorrow.

    Thanks!
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  2. #2
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    Quote Originally Posted by Latszer View Post
    If f(x)=1/x^2 find f'(x) using the definition of the derivative.
    (note: If you instead use the power rule, f'(x)=nx^(n-1), you will get zero credit for this problem. It is useful check of your answer though.)

    The answer is -2/x^3 and the best I can get is -1/x^3.
    Could someone get me on the right track? If you can't help by tomorrow, don't bother because it is due tomorrow.

    Thanks!
    f(x) = \frac{1}{x^2}


    f(x + h) = \frac{1}{(x + h)^2}


    f(x + h) - f(x) = \frac{1}{(x + h)^2} - \frac{1}{x^2}

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

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

     = \frac{x^2 - x^2 - 2xh - h^2}{x^2(x + h)^2}

     = \frac{h(-2x - h)}{x^2(x + h)^2}.


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

     = \frac{-2x - h}{x^2(x + h)^2}.


    \lim_{h \to 0}\frac{f(x + h) - f(x)}{h} = \lim_{h \to 0}\frac{-2x - h}{x^2(x + h)^2}

     = \frac{-2x}{x^2(x^2)}

     = -\frac{2x}{x^4}

     = -\frac{2}{x^3}.



    Thus f'(x) = -\frac{2}{x^3}.
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  3. #3
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    Thanks man, I factored wrong the first time I did it, I understand now.
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