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Math Help - Differential

  1. #1
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    Differential

    I was just wanting to know if some could show me how to work one of these:

    y = 3 x^2 + 3 x + 3

    Find the differential dy when x = 3 and dx = 0.4

    thus

    y' = 6x+3

    If someone could just one out I could pick on the pattern and finish the rest of them.

    Thanks!
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by qbkr21 View Post
    I was just wanting to know if some could show me how to work one of these:

    y = 3 x^2 + 3 x + 3

    Find the differential dy when x = 3 and dx = 0.4

    thus

    y' = 6x+3

    If someone could just one out I could pick on the pattern and finish the rest of them.

    Thanks!
    Another format for the derivative is:
    y' = dy/dx, so:
    dy/dx = 6x + 3

    Now "multiply" both sides by dx:
    dy = (6x + 3)dx

    Now just plug the numbers in.

    -Dan

    PS The "multiplication" I did really isn't a multiplication. It's a looooong story. Suffice it to say we can get away with it, but have a care to check the truth of it before you do it.
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    Re:

    What about if they ask you for change...

    Example:

    If y = 5sqrt(x)

    Find the change in Y, delta y when x = 1 and delta x = 0.3

    I could easily solve this problem if they gave me an additional Y coordinate. I tried looking in the textbook but it gives me some pretty bad examples. They use letters instead of #'s which to me adds to the confusion. Thanks for you previous help!

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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by qbkr21 View Post
    What about if they ask you for change...

    Example:

    If y = 5sqrt(x)

    Find the change in Y, delta y when x = 1 and delta x = 0.3

    I could easily solve this problem if they gave me an additional Y coordinate. I tried looking in the textbook but it gives me some pretty bad examples. They use letters instead of #'s which to me adds to the confusion. Thanks for you previous help!

    The "change in y" is usually written as (Delta)y. In this kind of problem we are approximating the derivative dy/dx as (Delta)y/(Delta)x. So it's the same kind of problem as before:
    dy = 5*(1/2)*1/sqrt(x) dx

    (Delta)y = 5*(1/2)*1/sqrt(x) * (Delta)x

    -Dan
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    Re:

    Dan You Are By God The Man!!!!!!!!!!!
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by qbkr21 View Post
    Dan You Are By God The Man!!!!!!!!!!!
    (I'm blushing!) You're welcome!

    -Dan
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    Quote Originally Posted by topsquark View Post
    The "change in y" is usually written as (Delta)y. In this kind of problem we are approximating the derivative dy/dx as (Delta)y/(Delta)x. So it's the same kind of problem as before:
    dy = 5*(1/2)*1/sqrt(x) dx

    (Delta)y = 5*(1/2)*1/sqrt(x) * (Delta)x

    -Dan

    However when I am working this problem it gives me the same answer we got before .750. I think that the answer must be something else...
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by qbkr21 View Post
    I think that the answer must be something else...
    why? does your text give a different answer?
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    Re:

    Webworks the online homework software does:

    Here are the two questions maybe this will make it a bit clear. Thank You for your help!


    Here are the 2 Problems (they are totally separate):

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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by qbkr21 View Post
    Webworks the online homework software does:

    Here are the two questions maybe this will make it a bit clear. Thank You for your help!


    Here are the 2 Problems (they are totally separate):
    ok, so i don't see why our method didn't work for the second one. i'll think about it. but lets work on the first.

    y = ln{sqrt[(6x + 5)/(7x - 4)]} ..........let's simplify this monstrosity a bit

    => y = (1/2)ln[(6x + 5)/(7x - 4)]
    => y = (1/2)ln(6x + 5) - (1/2)ln(7x - 4) ...........now let's differentiate
    => dy/dx = 3/(6x + 5) - (7/2)/(7x - 4)

    hold on, what did you want to do with this problem?
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    Quote Originally Posted by topsquark View Post
    The "change in y" is usually written as (Delta)y. In this kind of problem we are approximating the derivative dy/dx as (Delta)y/(Delta)x. So it's the same kind of problem as before:
    dy = 5*(1/2)*1/sqrt(x) dx

    (Delta)y = 5*(1/2)*1/sqrt(x) * (Delta)x

    -Dan
    Quote Originally Posted by qbkr21 View Post
    However when I am working this problem it gives me the same answer we got before .750. I think that the answer must be something else...
    x = 1, (Delta)x = 0.3

    (Delta)y = 5*(1/2)*1/sqrt(1)*0.3 = 0.75

    is what I get. I can't see what else it could be.

    -Dan
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    Re:

    I mean the big LN out in front confuses me I know that:


    (1/2)ln


    but then to I recopy the problem and log the 2nd half this, and this is the point in which I get confused. If the big LN were not there I would

    (1/2)[(6x+5)/(7x-4)^(-1/2) then I would differentiate the inside using the quotient rule for the inside of the square root. I really am stuck Jhevon.
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    Re:

    Here is the full problem this is just what it looks like on the computer:

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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by qbkr21 View Post
    What about if they ask you for change...

    Example:

    If y = 5sqrt(x)

    Find the change in Y, delta y when x = 1 and delta x = 0.3

    I could easily solve this problem if they gave me an additional Y coordinate. I tried looking in the textbook but it gives me some pretty bad examples. They use letters instead of #'s which to me adds to the confusion. Thanks for you previous help!

    see if this solution works:

    y = 5sqrt(x)

    now, (delta)y = f(x + (delta)x) - f(x)

    now f(x + deltax) = f(1 + 0.3) = f(1.3) = 5.700877125

    f(x) = f(1) = 5

    => deltay = 5.700877125 - 5 = 0.700877125
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  15. #15
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by qbkr21 View Post
    I mean the big LN out in front confuses me I know that:


    (1/2)ln


    but then to I recopy the problem and log the 2nd half this, and this is the point in which I get confused. If the big LN were not there I would

    (1/2)[(6x+5)/(7x-4)^(-1/2) then I would differentiate the inside using the quotient rule for the inside of the square root. I really am stuck Jhevon.
    You know what the ln function is yes? Well d/dx(ln(x)) = 1/x. Then you can use the chain rule from there, just like Jhevon did.

    -Dan
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