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anonymous

  • 5 years ago

the derivative of 2x^2 I need to know how to solve this using the derivative as a function... (using limits)

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  1. amistre64
    • 5 years ago
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    \[\lim_{h->0}\frac{f(x+h)-f(x)}{h}\]

  2. anonymous
    • 5 years ago
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    I already know that formula, i need to know how to use it....

  3. amistre64
    • 5 years ago
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    \[\lim\frac{2(x+h)^2-2x^2}{h}\] \[\lim\frac{2(x^2+2xh+h^2)-2x^2}{h}\] \[\lim\frac{2x^2+4xh+2h^2-2x^2}{h}\] \[\lim\frac{\cancel{2x^2}+4xh+2h^2\cancel{-2x^2}}{h}\] \[\lim\frac{4xh}{h}+\frac{2h^2}{h}\] \[\lim\frac{4x}{1}+\frac{2h}{1}\]

  4. amistre64
    • 5 years ago
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    if you know the formula, you know how to use it :/

  5. amistre64
    • 5 years ago
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    now, when h=0; 4x+2h = 4x

  6. amistre64
    • 5 years ago
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    when I first did these the "x" part of the function confused me a bit since they used x as part of the substituions

  7. amistre64
    • 5 years ago
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    f(x+h) should be written better as say: f(a+h) maybe so that you know to replace "x" by whats in the ( )

  8. anonymous
    • 5 years ago
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    ok, now i get it thanks alot

  9. amistre64
    • 5 years ago
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    youre welcome, and good luck :)

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