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Math Help - Differentiation using limits

  1. #1
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    Differentiation using limits

    HI there guys, I'm taking calculus for the third time and its been a long time since i can remember my basic algebra and such. So please I ask for patience for this simple work, I understand how to derive this:

    lim h->0 f(x+h)-f(x)/h; for example if f(x)=3/2x^2 i know it has to be written as:

    lim h->0 3/2(x+h)^2-3/2x^2/h

    What i'm having a hard time is including a i binomial and polynomial:

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

    All i ask is how to write it, I can figure the answer myself. So far I think its like this:

    lim h->0 2(x+3+h)-2x+3/h; however I'm not confident in that answer because all the x variables will be gone and with h=0 the answer be nothing. Please correct me where I am interpreting this wrong.
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  2. #2
    MHF Contributor Calculus26's Avatar
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    All i ask is how to write it, I can figure the answer myself. So far I think its like this:

    lim h->0 2(x+3+h)-2x+3/h; however I'm not confident in that answer because all the x variables will be gone and with h=0 the answer be nothing. Please correct me where I am interpreting this wrong.

    If f(x) =2x+ 3 then f(x+h) = 2(x+h) + 3 not 2(x+3+h)

    you obtain


    lim (2h/h) = lim2 = 2 be thankful for how easy it simplified

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

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

    and now you wish it were as simple as the first

    lim { [ 2(x+h)^2 + 3(x+h) -2] -[ 2x^2+3x-2]}/h
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    Thanks so much for this! Man this is going to be a long semester again, and I'm lucky i got a teacher this time only having us do the most basic problems.

    Honestly there were no examples in the book like this, thanks again.
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  4. #4
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    Quote Originally Posted by Calculus26 View Post
    If f(x) =2x+ 3 then f(x+h) = 2(x+h) + 3 not 2(x+3+h)

    you obtain


    lim (2h/h) = lim2 = 2 be thankful for how easy it simplified

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

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

    and now you wish it were as simple as the first

    lim { [ 2(x+h)^2 + 3(x+h) -2] -[ 2x^2+3x-2]}/h
    Since the first one is just 2, then when doing f'(-2) it can't be done correct? Since there are no x variables to place the -2, so its safe to say the problem is just done?
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  5. #5
    MHF Contributor Calculus26's Avatar
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    No just the opposite f' (x) = 2 for all x in particular f ' (-2) = 2

    What the Calculation shows is the the derivative is independent of x.

    This makes sense as 2x +3 is a linear function whose slope is 2 at every point

    It might be instriuctive for you to find f ' (x) using the limit definition when f(x) = mx + b

    not surprisingly again the x's cancel and you get m
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  6. #6
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    Quote Originally Posted by Calculus26 View Post
    No just the opposite f' (x) = 2 for all x in particular f ' (-2) = 2

    What the Calculation shows is the the derivative is independent of x.

    This makes sense as 2x +3 is a linear function whose slope is 2 at every point

    It might be instriuctive for you to find f ' (x) using the limit definition when f(x) = mx + b

    not surprisingly again the x's cancel and you get m

    Ahh ok i think i can picture in my head, if it were graph it then no matter at what x value it would be pretty much a straight line going threw 2 on the y axis correct? part of my hw asks to do f'(-2) f'(0) and f'(1) which would all be 2. I'm understanding now, my teacher isn't really asking for graphs so thanks!
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  7. #7
    MHF Contributor Calculus26's Avatar
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    If by "it " you mean the derivative as a function yes it is the horizontal line y =2
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    lim { [ 2(x+h)^2 + 3(x+h) -2] -[ 2x^2+3x-2]}/h

    i want to make sure i simplified this correctly:

    =2(x^2)+2xh+h^2)+3xh-2-2x^2+3x-2/h
    =2x^2+4xh+2h^2+3xh-2-2x^2+3x-2/h
    =7xh+2h^2+3x-4/h
    =lim h->0 7x+2h+3x-4
    =10x-4

    right?
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  9. #9
    MHF Contributor Calculus26's Avatar
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    lim { [ 2(x+h)^2 + 3(x+h) -2] -[ 2x^2+3x-2]}/h
    is correct to here but

    =2(x^2)+2xh+h^2)+3xh-2-2x^2+3x-2/h


    should be [2x^2 + 4xh + 2h^2 +3x +3h -2 - 2x^2-3x +2]/h


    = 4x + 3 + 2h

    = 4x +3 after taking limit
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  10. #10
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    just a double check on using x^3

    =(x+h)^3-x^3
    =x^3+3x^2h+3xh^2+h^3-x^3/h
    =3x^2h+3xh^2+h^3/h
    =3x^2+3xh+h^2
    =3x^2+3x

    The final part is what makes me unsure, would the h->0 next to the 3x leave it untouched like i think or cancel it out as it may be multiplied?
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  11. #11
    MHF Contributor Calculus26's Avatar
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    you're ok to here:

    3x^2+3xh+h^2

    Now lim 3x^2+3xh+h^2 = 3x^2
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  12. #12
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     \lim_{h \to 0} \frac{x^3+3x^2h+3xh^2+h^3-x^3}{h}

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

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

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

      3x^2+3x(0)+0^2

      3x^2
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  13. #13
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    Thank you for the visualization, i guess its just best to see the 3 and 0 multiply and become zero, i just dont understand why the x variable is canceled out as well.
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  14. #14
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    Something to remember...

     0 \times \text{Anything!} = 0

    i.e.

    0\times 1 = 0

    0\times 2 = 0

    0\times 3 = 0

    0\times x = 0

    0\times y = 0

    0\times 3x = 0

    0\times 3xy = 0
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