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Math Help - What does dy/dx REALLY mean?

  1. #1
    Member garymarkhov's Avatar
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    What does dy/dx REALLY mean?

    I'm trying to understand the intuition and precise meaning of each component of common differentiation notation.

    As I understand it:

    <br />
\frac{dy}{dx} = f'(x) = y'<br />

    Each form of notation asks "for our function of x, what is the slope at point x?" Or, in the case of <br />
\frac{dy}{dx}<br />
, "how much of 'little bit of y' is there for every 'little bit of x'?" (which is the same as asking what the slope is). So far so good. But suppose we have a basic function <br />
y = x^2<br />
. What is the correct <br />
\frac{dy}{dx}<br />
notation for this derivative, and what is the specific meaning of each part?

    Is it:

    <br />
\frac{dy}{dx}y=\frac{dy}{dx}x^2<br />

    The right side makes sense ("how much change in y is there for every change in x when we consider the function <br />
x^2<br />
?"), but the left side doesn't seem to. Can you "ask" <br />
\frac{dy}{dx}<br />
of y? How do we get just <br />
\frac{dy}{dx}<br />
instead of <br />
\frac{dy}{dx}y<br />
?

    Alternatively, we could do:

    <br />
\frac{d}{dx}y=\frac{d}{dx}x^2<br />

    Now the left side makes sense. We want to solve for <br />
\frac{dy}{dx}<br />
, and placing just a <br />
\frac{d}{dx}<br />
next to the y allowed the y that was already there to complete the notation. But the right side doesn't make sense anymore. What is the numerator "asking"? If we allowed x to complete the question like we did on the left side, we would get something like <br />
\frac{dy}{dx}=\frac{dx}{dx}x<br />
, but that is almost certainly incorrect.

    Or maybe I didn't get it right with either option I offered up. Let me know. And whatever it is, please specify what the notation '"is asking" in each component.
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    You have a very good understanding of this. \frac{dy}{dx} is very different from \frac{d}{dx} though. The difference is the first expression is the same as f'(x) or y' and means "the derivative of function y with respect to x". It is not an operation on anything. The second expression I mentioned though is just that. It means take the derivative of whatever comes after d/dx with respect to x.

    So again, the difference is between saying "this is the derivative of y" and "take the derivative of what's after this with respect to x".
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    when you take the derivative using the d/dx notation it's always:

    \frac{d}{dx}( f(x) )

    So where you have:

    y=x^2

    you want the derivative you do:

    \frac{d}{dx}(y)

    substituting by x^2 for y, by the equation above, you get:

    =\frac{d}{dx}(x^2)

    by power rule: =2x\frac{dx}{dx}

    so you divide: =2x*1

    so in the end:

    \frac{d}{dx}(y)=\frac{dy}{dx}=2x

    However those are just symbols. The truth is that the derivative is:

    \lim_{\Delta x \rightarrow 0}\frac{\Delta y}{\Delta x}=\frac{dy}{dx}

    That may seem like more symbols, what it means is say I'm looking at the slope. Only that the "lim" thing is asking what happens when delta X goes to 0? It just means that you're shrinking the difference between points you're looking at, until they become the same point.

    so like y=2x.

    you know the slope is 2.

    well how do you calculate it? you choose 2 points, say: (1,2) and (5,10).

    so then you take: \frac{\Delta y}{\Delta x}=\frac{y_2-y_1}{x_2-x_1}

    you plug in the 2 points:

    \frac{10-2}{5-1}=\frac{8}{4}=2

    Well so that's the slope, well what if we were to look at when delta x goes to 0?

    You then are saying that the difference between the X values are getting smaller and smaller until you're looking at the slope exactly at some point. In the case of my example, there is a constant slope.


    For something like y=x^2

    The slope is not constant. We however can look at the slope at different points, by drawing a tangent line at that point, the slope of the tangent line is the slope of the parabola at that point.

    so hence: \frac{dy}{dx}=2x.

    If you look at the point (1,1) on the parabola (if you have a graphing program or a relatively accurate drawing) and you make a line tangent to the point on the parabola, you'll see that the line has a slope of 2.


    I hope I didn't confuse you. All it really boils down to is that the derivative is the slope of a curve at any point on the curve.


    remark: It's also commonly defined as: f'(x)=\lim_{\Delta h \rightarrow 0}\frac{f(x+h)-f(x)}{h}

    and that's the same as: \lim_{\Delta x \rightarrow 0}\frac{\Delta  f(x)}{\Delta x}
    Last edited by seld; September 9th 2009 at 08:15 PM. Reason: fixing notation on the remark
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  4. #4
    Member garymarkhov's Avatar
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    Quote Originally Posted by Jameson View Post
    So again, the difference is between saying "this is the derivative of y" and "take the derivative of what's after this with respect to x".
    Perfect. Thank you for the fast and helpful response. It is good to know that I'm not altogether out of my mind.

    So for the function <br />
y = x^2, <br />
\frac{d}{dx}<br />
operates on <br />
x^2 and delivers <br />
\frac{dy}{dx}=2x<br />
. Good stuff.

    So now my question is why, if \frac{dy}{dx} is so different from \frac{d}{dx}, is it possible to snap components on and off so easily when doing implicit differentiation?

    Consider y^2+x^2 = 100

    When we differentiate with respect to x, we can write \frac{d}{dx}x^2 + \frac{d}{dx}y^2 = \frac{d}{dx}100

    And since we don't know how to calculate \frac{d}{dx}y^2, we differentiate that element with respect to y and then tack on (dare I say "apply"?) \frac{dy}{dx} after to get \frac{d}{dy}y^2\frac{dy}{dx}=2y\frac{dy}{dx}. And I have it stuck in my mind that we do that bit of dy cancellation to convince ourselves that what we've really found is \frac{dy}{dx}=\frac{d}{dx}y^2. Is that what's going on? Are we free to cancel components of two dissimilar notational items like that?

    Quote Originally Posted by seld View Post
    substituting by x^2 for y, by the equation above, you get:

    =\frac{d}{dx}(x^2)

    by power rule: =2x\frac{dx}{dx}
    Most of what you've written makes a great deal of sense. Thanks! However, I am confused about why you multiplied (is that the right word?) 2x by \frac{dx}{dx}. I agree it probably won't do anybody any harm to tack on \frac{dx}{dx} to the right of 2x, but is that actually going on when you compute \frac{d}{dx}(x^2)? Is it a real step that gets omitted because it's so obvious?
    Last edited by garymarkhov; September 9th 2009 at 11:27 PM. Reason: Eliminated "=" signs from inappropriate places; fixed improper differentiation of x^2
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    Quote Originally Posted by garymarkhov View Post
    Perfect. Thank you for the fast and helpful response. It is good to know that I'm not altogether out of my mind.

    So for the function <br />
y = x^2, <br />
\frac{d}{dx}<br />
operates on <br />
x^2 and delivers <br />
\frac{dy}{dx}=x^2<br />
. Good stuff.
    There's an error there, the derivative of x^2 != x^2.

    derivative of x^2 = 2x.

    Quote Originally Posted by garymarkhov View Post
    So now my question is why, if \frac{dy}{dx} is so different from \frac{d}{dx}, is it possible to snap components on and off so easily when doing implicit differentiation?

    Consider y^2+x^2 = 100

    When we differentiate with respect to x, we can write \frac{d}{dx}x^2 + \frac{d}{dx}y^2 = \frac{d}{dx}100

    And since we don't know how to calculate \frac{d}{dx}y^2, we differentiate that element with respect to y and then tack on (dare I say "apply"?) \frac{dy}{dx} after to get \frac{d}{dy}y^2\frac{dy}{dx}=2y\frac{dy}{dx}. And I have it stuck in my mind that we do that bit of dy cancellation to convince ourselves that what we've really found is \frac{dy}{dx}=\frac{d}{dx}y^2. Is that what's going on? Are we free to cancel components of two dissimilar notational items like that?



    However, I am confused about why you multiplied (is that the right word?) 2x by \frac{dx}{dx}. I agree it probably won't do anybody any harm to tack on \frac{dx}{dx} to the right of 2x, but is that actually going on when you compute \frac{d}{dx}(x^2)? Is it a real step that gets omitted because it's so obvious?
    These last 2 questions are somewhat related.

    As you saw earlier, I wrote it with the dx/dx intentionally. You always have to remember what your differentiating by.

    To take your circle:

    x^2+y^2=100

    you want to take the derivative of it, which means you want to look at how y changes in terms of x because you're still on the xy cartesian plane (aka you want dy/dx).

    So you do:
    \frac{d}{dx}(x^2+y^2)=\frac{d}{dx}(100)

    then that equals:
    2x*\frac{dx}{dx}+2y*\frac{dy}{dx}=0

    Derivatives of constants = 0 because you're looking at the how it's changing as x changes, but constants don't change, so it's 0.

    You asked me why did I include dx/dx because normally it's omitted, but it makes a difference when you do implicit differentiation.

    You see the dy/dx term that I included? it's the same as the dx/dx the only difference is that it's a y. There's a reason for this, it's because when you differentiate something in terms of x by x, you're looking at how x changes in terms of x. Whereas looking at how y changes in terms of x is fundamentally different.

    back to the example:

    You're looking for dy/dx however you have some junk + a dy/dx = 0.

    So you shove things around.
    2y*\frac{dy}{dx}=-2x
    \frac{dy}{dx}=\frac{-2x}{2y}
    \frac{dy}{dx}=-\frac{x}{y}

    You can convince yourself that that is the solution when you start drawing tangent lines on your circle and look at the points. So like the point, (10,0), yields an undefined slope. But if you think about that, (10,0) is the rightmost point on your circle, if you draw a tangent line to that, it's a vertical line. A vertical line has an undefined slope because well, y is going from (-\infty,\infty). However x is not changing at all.

    Also when you do chain rule, it becomes clear as well:

    in prime notation:
    (f(x) \dot  g(x))'=f'((g(x))*g'(x)

    in d/dx notation:
    \frac{dy}{dx}=\frac{dy}{du}*\frac{du}{dx}

    (i'll let you convince yourself that those are the same)


    y=x^{x^3}

    \frac{d}{dx}(x^{x^3})


    \text{Let } u = x^3

    part 1:

    \frac{dy}{du}=\frac{d}{du}(x^{u})

    power rule:
    =ux^{u-1}*\frac{du}{du}

    =ux^{u-1}
    Then you do the other part:

    \frac{d}{dx}(u)=\frac{d}{dx}(x^3)

    =3x^2*\frac{dx}{dx}

    so combine the 2:

    \frac{dy}{du}*\frac{du}{dx}

    ux^{u-1}*3x^2

    Then you plug in x^3 for u.

    x^3x^{x^3-1}*3x^2

    I'll let you simplify it.


    Hope that helps.
    Last edited by seld; September 10th 2009 at 10:40 AM. Reason: fixing grammar
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  6. #6
    Member garymarkhov's Avatar
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    Quote Originally Posted by seld View Post
    There's an error there, the derivative of x^2 != x^2.
    Yowza. Thanks.

    As you saw earlier, I wrote it with the dx/dx intentionally. You always have to remember what your differentiating by.
    Sure, but what does it mean to multiply your result by \frac{dx}{dx}. Are you doing so as a memory aid? I'm looking for the real meaning here. I can see why we would write \frac{d}{dx}f(x)=\frac{dy}{dx}, but I fail to see why we would write f'(x)\frac{dx}{dx}=\frac{dy}{dx}.

    You see the dy/dx term that I included? it's the same as the dx/dx the only difference is that it's a y. There's a reason for this, it's because when you differentiate something in terms of x by x, you're looking at how x changes in terms of x. Whereas looking at how y changes in terms of x is fundamentally different.
    I agree, but my question above remains.

    You can convince yourself that that is the solution when you start drawing tangent lines on your circle and look at the points.
    I've long been geometrically-convinced and algebraically-skeptical.

    Also when you do chain rule, it becomes clear as well:

    in prime notation:
    (f(x)\dot g(x))'=f'((g(x))*g'(x)
    is the "dot" g intentional? I would have written (f(x)g(x))'=f'(x)g(x)+f(x)g'x). Is that the same?

    in d/dx notation:
    \frac{dy}{dx}=\frac{dy}{du}*\frac{du}{dx}

    (i'll let you convince yourself that those are the same)
    Deep in my heart (and within my thick skull), I can't. That's my problem!


    \frac{dy}{du}=\frac{d}{du}(x^{u})
    Aren't we taking the derivative with respect to x here? It won't affect the results, but I'm mostly interested in the notation (as I'm sure you've gathered by now from my incessant yammering about it).

    -------------------------------

    I think if you can clear this up, almost all of the rest will go away:

    Regarding the \frac{d}{dx}y^2 component in my circle.

    We can say that \frac{d}{dx}y^2=\frac{dy}{dx}, right? Can we also say that \frac{d}{dy}y^2=\frac{dy}{dy}? But 2y != 1, so I must be going wrong. And yet it seems the only route to getting to write 2y\frac{dy}{dx}.
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    [COLOR=Black]There's a reason for the dot, it's because I don't know how to get latex to do a f * g composite function thingie. I know there's a way to get a dot, I didn't realize it would put it over the g like that.


    So it doesnt' really mean anything when you multiply by dx/dx, I wrote it like that to be clear when you have like a composite function or you're differentiating implicitly. From what I remember in calculus there are some details that go on and ultimately it means dx/dx =1.


    It's also related to the your question about the chain rule.



    "algebraically" you can say \frac{dy}{dx}=\frac{dy}{du}*\frac{du}{dx}=\frac{dy  }{dx}*\frac{du}{du}

    Though this is technically incorrect, however it is more or less what happens.



    And no, when it says d/du you're taking the derivative with respect to u. That's what you're doing when you do substitution.


    Going back to the implicit differentiation

    \frac{d}{dx}y^2=\frac{dy}{dx}

    No, because if you think about it, when you have
    \frac{d}{dx}(y)=\frac{dy}{dx}

    right? well you have the same situation with x.
    \frac{d}{dx}(x)=\frac{dx}{dx}=1

    does that make sense?
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    Member garymarkhov's Avatar
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    So it doesnt' really mean anything when you multiply by dx/dx, I wrote it like that to be clear when you have like a composite function or you're differentiating implicitly. From what I remember in calculus there are some details that go on and ultimately it means dx/dx =1.
    I very much want to know what those details are. If anyone knows, please share.

    "algebraically" you can say \frac{dy}{dx}=\frac{dy}{du}*\frac{du}{dx}=\frac{dy  }{dx}*\frac{du}{du}

    Though this is technically incorrect, however it is more or less what happens.
    I'm sorry, but I've always been frustrated by explanations of this type. If you can point me to a resource that provides a more thorough treatment of this, that would be great.

    No, because if you think about it, when you have
    \frac{d}{dx}(y)=\frac{dy}{dx}

    right? well you have the same situation with x.
    \frac{d}{dx}(x)=\frac{dx}{dx}=1

    does that make sense?
    No. In Post #4 we established that for a function y=x^2, \frac{d}{dx}(y)=\frac{d}{dx}(x^2)=\frac{dy}{dx}.
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    Quote Originally Posted by garymarkhov View Post
    I very much want to know what those details are. If anyone knows, please share.

    I'm sorry, but I've always been frustrated by explanations of this type. If you can point me to a resource that provides a more thorough treatment of this, that would be great.
    Though that explanation is used commonly because you can relate it to fractions and then people tend to get the idea you can cancel stuff. They essentially tell you to treat the dy/dx notation as fractions. I'm sorry I can't help with either, it's been so long since I've been in a calculus class. I suppose you could use the base definition of a limit.

    Limit Def:

    Let f(x) be a function defined on an interval that contains x=a, except possibly at x=a. Then we say that,

    \lim_{x \rightarrow a}f(x)=L

    if for all \epsilon > 0 there exists a number \delta > 0 such that

    |f(x)-L|<\epsilon \text{   whenever    } 0<|x-a|<\delta


    And use that to evaluate the definition of the derivative:

    f'(x)=\lim_{\Delta h \rightarrow 0}\frac{f(x+h)-f(x)}{h}<br />



    Then the details should surface, and if you do it to the dy/du*du/dx thing it could give you some illumination as well. Though the definition of the limit basically comes down to saying is that the limit is L if you can find a delta where

    0<|x-a|<delta is true, and |f(x)-L| < epsilon.

    Normally you do it by picking a delta after you've plugged in the L and stuff (though I vaguely remember the process)



    Quote Originally Posted by garymarkhov View Post
    No. In Post #4 we established that for a function y=x^2, \frac{d}{dx}(y)=\frac{d}{dx}(x^2)=\frac{dy}{dx}.

    Yes what you're saying is right, the derivative of y = dy/dx.
    but I did not define a y when I posted:
    \frac{d}{dx}(y)=\frac{dy}{dx}<br />

    I was referring to a general case, not a specific case and what you have isn't wrong. It pertains to a specific case of when y=x^2, and that's where you get the middle step. If you had no idea what y was, you wouldn't have the center bit. The idea being if you have some function y, the derivative of y is the derivative of y.
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    Thanks for all your help Seld. You've helped clarify quite a bit.

    Still, I think we're both in the same boat in not having a deep understanding of how we go from

    \frac{d}{dx}x^2 + \frac{d}{dx}y^2 = \frac{d}{dx}100<br />

    to

    2x*\frac{dx}{dx}+2y*\frac{dy}{dx}=0

    besides that "it works" and "it's convention". Does anyone else want to chime in with an intuitive interpretation of what's going on here? How do we justify slipping \frac{dx}{dx} and \frac{dy}{dx} in there?
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    The notation that we use is arbitrary and just must be accepted. Try not to think of derivative notations as fractions because you cannot apply the same rules you do with normal fractions.

    As far as justifying the derivatives you mentioned, you need to go back to the limit definition of a derivative. From this definition all of the standard rules (power rule, quotient rule, etc.) can be derived. It really depends on what level of rigorousness you want to use.

    I guess my main point is to not get in the habit of thinking that 2x(dx/dx)=2x because the dx's cancel out. That's not the way the definitions of derivatives were constructed or proved.
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    Quote Originally Posted by Jameson View Post
    The notation that we use is arbitrary and just must be accepted. Try not to think of derivative notations as fractions because you cannot apply the same rules you do with normal fractions.

    As far as justifying the derivatives you mentioned, you need to go back to the limit definition of a derivative. From this definition all of the standard rules (power rule, quotient rule, etc.) can be derived. It really depends on what level of rigorousness you want to use.

    I guess my main point is to not get in the habit of thinking that 2x(dx/dx)=2x because the dx's cancel out. That's not the way the definitions of derivatives were constructed or proved.

    true, sorry if I wasn't being clear, I meant that they liken it to fractions so that people can find them more palatable. I didn't mean that treating


    \frac{dy}{dx}=\frac{dy}{du}*\frac{du}{dx}=\frac{dy  }{dx}*\frac{du}{du}

    like a fraction and commuting things around isn't rigorously true. Sorry if there was a misunderstanding when I "spoke" so to speak. I merely meant that that's the explanation most commonly used (at least that was the excuse my teachers gave).
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    Try not to think of derivative notations as fractions because you cannot apply the same rules you do with normal fractions.
    That's an important insight. Thank you.

    As far as justifying the derivatives you mentioned, you need to go back to the limit definition of a derivative. From this definition all of the standard rules (power rule, quotient rule, etc.) can be derived. It really depends on what level of rigorousness you want to use
    I am satisfied that the derivatives this method serves up are correct. No issues there.

    The notation that we use is arbitrary and just must be accepted
    I am reasonably certain that the first person who decided to use this notation didn't do so because it magically produced the right answer even though he couldn't think of a logical reason why it should. Some notation can afford to be arbitrary, but not when it works its way into an equation and affects the outcome.

    I know I'm being tremendously boneheaded about this, and I apologize for that. I want to confirm that I greatly appreciate your help so far. But I cannot abide "The notation that we use is arbitrary and just must be accepted". We all know what it was like to be told "just deal with it - it works" by some teachers, only to discover later that there was a real and insightful answer - they just weren't capable of providing it.

    So explain it, point me to some resource that can explain it, or say, "I don't know". But I sincerely hope we can avoid ending this with an admonition to "just accept it", as well meaning as it was. Math is too fun and too important!
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    Quote Originally Posted by garymarkhov View Post
    ... I cannot abide "The notation that we use is arbitrary and just must be accepted". We all know what it was like to be told "just deal with it - it works" by some teachers, only to discover later that there was a real and insightful answer - they just weren't capable of providing it.

    So explain it, point me to some resource that can explain it, or say, "I don't know". But I sincerely hope we can avoid ending this with an admonition to "just accept it", as well meaning as it was. Math is too fun and too important!
    When you do get to a point of enlightenment and work out exactly what it *does* all mean, make a point of writing this down and publishing it - there's money to be made in explaining this sort of concept, and it's rarely done well
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    @ Matt: If I ever get this nailed down, I'll be sure to report back

    @ Jameson: I recently came across two resources that treat \frac{dy}{dx} notation like a fraction.

    First example: The Most Beautiful Equation in the World The Math Less Traveled. Getting from \frac{d\theta}{dz}=iz to \int{z}dz = \int{\theta}d\theta

    Second example: . The question begins at the beginning of the video, but our point of interest is just after 08:30 --> getting from \frac{\mu(x)}{x}=\frac{du}{dx} to \frac{1}{x}dx=\frac{1}{\mu(x)}dx

    In both of these scenarios, the pieces of what you said should not be treated as a fraction are being treated as if they are. Is this usage incorrect, but happens to work in these scenarios? In what scenarios would this usage fail to work? Do you stand by your advice to "Try not to think of derivative notations as fractions because you cannot apply the same rules you do with normal fractions"?
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