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Math Help - Derivatives with x and h variables

  1. #1
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    Derivatives with x and h variables

    Limit of h goes to 0 [2(x+h)^2+5(x+h)^4]-[2x^2+5x^4]/h=?
    (Hint: This is the definition of the derivative.)

    Obviously the hint is giving me a big clue, but I don't understand the concept. My limits are kinda rusty, and I'm not sure what to do. Help Please.
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  2. #2
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    Quote Originally Posted by DarkestEvil View Post
    Limit of h goes to 0 [2(x+h)^2+5(x+h)^4]-[2x^2+5x^4]/h=?
    (Hint: This is the definition of the derivative.)

    Obviously the hint is giving me a big clue, but I don't understand the concept. My limits are kinda rusty, and I'm not sure what to do. Help Please.
    you don't need to "work out" the limit ... all you have to do is remember that

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

    ... you have f(x).
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  3. #3
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    Quote Originally Posted by DarkestEvil View Post
    Limit of h goes to 0 [2(x+h)^2+5(x+h)^4]-[2x^2+5x^4]/h=?
    (Hint: This is the definition of the derivative.)

    Obviously the hint is giving me a big clue, but I don't understand the concept. My limits are kinda rusty, and I'm not sure what to do. Help Please.
    Just expand and do the algebra.
    (x+h)^2=x^2+2xh+h^2

    (x+h)^4=x^4+4x^3h+6x^2h^2+4xh^3+h^4
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  4. #4
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by DarkestEvil View Post
    Limit of h goes to 0 [2(x+h)^2+5(x+h)^4]-[2x^2+5x^4]/h=?
    (Hint: This is the definition of the derivative.)

    Obviously the hint is giving me a big clue, but I don't understand the concept. My limits are kinda rusty, and I'm not sure what to do. Help Please.
    f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}

    So in this example, f(x)=2x^2+5x^4 and therefore, using the power rule, the limit equals f'(x)=4x+20x^3.
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