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Thread: dy/dx = dy/du * du/dx

  1. #1

    dy/dx = dy/du * du/dx

    Hey.

    I was learning about how to differentiate composite functions. This is what I read:

    "If y = c(ax + b) + d.
    let u = ax + b, then y = cu + d.
    So dy/du = c, du/dx = a, and dy/dx = ca, that is dy/dx = dy/du * du/dx

    The equation dy/du = c means that y is changing c times as fast as u; similarly u is changing a times as fast as x. It is natural to think that if y is changing c times as fast as u, and u is changing a times as fast as x, then y is changing c*a times as fast as x. Thus again, dy/dx = dy/du * du/dx"

    I'm not surprised that it wasn't natural for me to think that way, and even after thinking about it, I still don't understand. I just can't get my head around it! Could somebody explain

  2. #2
    Elite Member
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    Quote Originally Posted by burgerandcheese View Post
    Hey.

    I was learning about how to differentiate composite functions. This is what I read:

    "If y = c(ax + b) + d.
    let u = ax + b, then y = cu + d.
    So dy/du = c, du/dx = a, and dy/dx = ca, that is dy/dx = dy/du * du/dx

    The equation dy/du = c means that y is changing c times as fast as u; similarly u is changing a times as fast as x. It is natural to think that if y is changing c times as fast as u, and u is changing a times as fast as x, then y is changing c*a times as fast as x. Thus again, dy/dx = dy/du * du/dx"

    I'm not surprised that it wasn't natural for me to think that way, and even after thinking about it, I still don't understand. I just can't get my head around it! Could somebody explain
    y = c(ax + b) + d = cax+cb+d. So dy/dx =ca
    A mathematician is a blind man in a dark room looking for a black cat which isnít there. - Charles R. Darwin

  3. #3
    Quote Originally Posted by Jomo View Post
    y = c(ax + b) + d = cax+cb+d. So dy/dx =ca
    Well, yes of course! That's the easier way

  4. #4
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    Quote Originally Posted by burgerandcheese View Post
    Hey.

    I was learning about how to differentiate composite functions. This is what I read:

    "If y = c(ax + b) + d.
    let u = ax + b, then y = cu + d.
    So dy/du = c, du/dx = a, and dy/dx = ca, that is dy/dx = dy/du * du/dx

    The equation dy/du = c means that y is changing c times as fast as u; similarly u is changing a times as fast as x. It is natural to think that if y is changing c times as fast as u, and u is changing a times as fast as x, then y is changing c*a times as fast as x. Thus again, dy/dx = dy/du * du/dx"

    I'm not surprised that it wasn't natural for me to think that way, and even after thinking about it, I still don't understand. I just can't get my head around it! Could somebody explain
    Suppose I run twice as fast as Tim, and Tim runs 3 times as fast as Dan. Then I run 6 times as fast as Dan.

    Is that the part you don't see, or is it how that relates to the derivative?

  5. #5
    Elite Member
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    Quote Originally Posted by burgerandcheese View Post
    Well, yes of course! That's the easier way
    That explains why the derivative is c*a
    A mathematician is a blind man in a dark room looking for a black cat which isnít there. - Charles R. Darwin

  6. #6
    Quote Originally Posted by Dr.Peterson View Post
    Suppose I run twice as fast as Tim, and Tim runs 3 times as fast as Dan. Then I run 6 times as fast as Dan.

    Is that the part you don't see, or is it how that relates to the derivative?
    Wow.. I get it now!!!!!!!! It was that simple.. why was I so dumb. No, not how it relates to the derivative.

    Thanks Dr.Peterson, and thanks Jomo!!

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