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Math Help - Implicit Differentiaton

  1. #1
    Bar0n janvdl's Avatar
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    Implicit Differentiaton

    Please help me with this guys.

    Could somebody please clarify for me what does it mean exactly, in simple terms?

    Then i have the following problem i have some questions about:

     y^5 + 3x^2 = sin x - 4y^3

    Then we differentiate...

     (5y^4 . y') + (6x) = (cos x) - (12y^2 . y')

    Ok, now where did that y' come from and what exactly does it mean?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by janvdl View Post
    Please help me with this guys.

    Could somebody please clarify for me what does it mean exactly, in simple terms?

    Then i have the following problem i have some questions about:

     y^5 + 3x^2 = sin x - 4y^3

    Then we differentiate...

     (5y^4 . y') + (6x) = (cos x) - (12y^2 . y')

    Ok, now where did that y' come from and what exactly does it mean?
    the y' is actually short for dy/dx

    i think i told you what this notation meant before, dy/dx means "the derivative of y with respect to x"

    in implicit differentiation we have to take account of what variable we are differentiating, and what variable we are differentiating with respect to. in the question you posted, you wanted to take the derivative with respect to x. so when we take the derivative of an x term, we attach dx/dx (we took "the derivative of x with respect to x"), however, since derivative notations can function as fractions, the dx's cancel and it becomes 1, so you don't see it, since it's like multiplying by one. however, when we take the derivative of a y term, we attach dy/dx (since we took "the derivative of y with respect to x"), and that doesn't cancel, so it stays. we write y' when it's understood what we are differentiating with respect to.

    we need implicit differentiation to differentiate when we have mixed terms that we can't really separate. it would be hard to solve for y and then find y' that way, so we do it implicitly
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    Bar0n janvdl's Avatar
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    Quote Originally Posted by Jhevon View Post
    the y' is actually short for dy/dx

    i think i told you what this notation meant before, dy/dx means "the derivative of y with respect to x"

    in implicit differentiation we have to take account of what variable we are differentiating, and what variable we are differentiating with respect to. in the question you posted, you wanted to take the derivative with respect to x. so when we take the derivative of an x term, we attach dx/dx (we took "the derivative of x with respect to x"), however, since derivative notations can function as fractions, the dx's cancel and it becomes 1, so you don't see it, since it's like multiplying by one. however, when we take the derivative of a y term, we attach dy/dx (since we took "the derivative of y with respect to x"), and that doesn't cancel, so it stays. we write y' when it's understood what we are differentiating with respect to.

    we need implicit differentiation to differentiate when we have mixed terms that we can't really separate. it would be hard to solve for y and then find y' that way, so we do it implicitly
    Ok, so every y in the term has to be multiplied with y' ?

    By the way Jhevon, u gota be online on msn before i can send you a message. Im using my phone for msn.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by janvdl View Post
    Ok, so every y in the term has to be multiplied with y' ?

    By the way Jhevon, u gota be online on msn before i can send you a message. Im using my phone for msn.
    yes. and if you were differentiating with respect to another variable besides x (which you will do when you start related rates) you would have to add x' to all the x terms as well
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    Bar0n janvdl's Avatar
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    Quote Originally Posted by Jhevon View Post
    which you will do when you start related rates...
    Thats next years worries Jhevon.
    Thanks for the help guys.
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by janvdl View Post
    Thats next years worries Jhevon.
    Thanks for the help guys.
    actually, related rates is usually the topic taught right after (or very soon after) implicit differentiation, so it's closer than you think
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  7. #7
    Bar0n janvdl's Avatar
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    Quote Originally Posted by Jhevon View Post
    actually, related rates is usually the topic taught right after (or very soon after) implicit differentiation, so it's closer than you think
    I'm actually not supposed to be doing this kind of stuff already.

    I'm only supposed to start with it next year.

    I do it for the fun, because it's a new challenge for me.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by janvdl View Post
    I'm actually not supposed to be doing this kind of stuff already.

    I'm only supposed to start with it next year.

    I do it for the fun, because it's a new challenge for me.
    i know, but there's no harm in getting a head start on related rates anyway. it proves a challenging topic for a lot of students, it would do you good to tackle it now
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