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Math Help - Derivative using a limit definition

  1. #1
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    Question Derivative using a limit definition

    It's pretty straight forward procedure.

    We have a function:

    f(x) = 2x^2 + 3x

    To get the derivative, we use the limit definition.

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

    \lim_{h\to\0}\frac{2(x+h)^2 + 3(x+h) - 2x^2 +3x}{h}

    \lim_{h\to\0}\frac{2x^2 + 2xh + h^2 + 3x + 3h  - 2x^2 +3x}{h}

    \lim_{h\to\0}\frac{4xh + 2h^2 +3h + 6x}{h}

    But how can I get rid of 6x in the numerator? Because that's the problem, I can't get the proper derivative, if the 6x is in the numerator.
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  2. #2
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    Re: Derivative using a limit definition

    Quote Originally Posted by Nforce View Post
    It's pretty straight forward procedure.

    We have a function:

    f(x) = 2x^2 + 3x

    To get the derivative, we use the limit definition.

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

    \lim_{h\to\0}\frac{2(x+h)^2 + 3(x+h) - 2x^2 +3x}{h}

    \lim_{h\to\0}\frac{2x^2 + 2xh + h^2 + 3x + 3h  - 2x^2 +3x}{h}

    \lim_{h\to\0}\frac{4xh + 2h^2 +3h + 6x}{h}

    But how can I get rid of 6x in the numerator? Because that's the problem, I can't get the proper derivative, if the 6x is in the numerator.
    second line :

    +3x should be -3x because -f(x)=-2x^2-3x
    Thanks from a tutor and Nforce
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  3. #3
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    Re: Derivative using a limit definition

    Beaten to it by princeps.
    Thanks from princeps
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  4. #4
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    Re: Derivative using a limit definition

    Oh, that thing I didn't saw. Thank you.
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  5. #5
    Member Sylvia104's Avatar
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    Re: Derivative using a limit definition

    In fact you made a few mistakes.

    Quote Originally Posted by Nforce View Post
    \lim_{h\to\0}\frac{f(x + h) - f(x)}{h}

    \lim_{h\to\0}\frac{2(x+h)^2 + 3(x+h) - 2x^2 \color{red}-\color{black}3x}{h}

    \lim_{h\to\0}\frac{2x^2 + \color{red}4\color{black}xh + \color{red}2\color{black}h^2 + 3x + 3h  - 2x^2 \color{red}-\color{black}3x}{h}
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  6. #6
    Member sbhatnagar's Avatar
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    Re: Derivative using a limit definition

    \begin{aligned} y =x^n \quad \text{then} \quad \frac{dy}{dx}&= \lim_{h \to 0} \frac{(x+h)^n-x^n}{h} \\ &=\lim_{h \to 0}\left[\frac{1}{h}\sum_{r=0}^{n}\binom{n}{r}x^{n-r}h^r-\frac{x^n}{h}\right] \\ &= \lim_{h \to 0}\left[ nx^{n-1}+\binom{n}{2}x^{n-2}h + \cdots +h^{n-1}\right] \\ &= nx^{n-1}\end{aligned}

    This fact can now be easily applied to the original problem.
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