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Leibniz notation when taking derivatives!

  1. Sep 15, 2012 #1

    I am encountering some major confusion. When taking just garden variety f(x)=y derivatives of the form dy/dx, I don't encounter any problems. But recently I started taking derivatives of parametric equations, or switching things up using polar equations and I realized perhaps I'm not so solid on these things as I once thought.

    Is there any way someone could tell me if I'm right about a few problems?

    So say we have:


    if we go (d/dx)(2x2+3x) the answer is just (4x+3)

    Now what if x is actually a function of another variable, say, t for instance. Say we know in reality, that x=2t.

    Now is (d/dx)(2x2+3x) still (4x+3)? after all, we are only asking about the differential in terms of x

    Now (d/dt)(2x2+3x) = (d/dx)(dx/dt)(2x2+3x) = (4x+3)(2)


    Ok so now what if we have




    Also, (dx/dt)=4, which is the same as writing (d/dt)(4t)=4

    Now, we could also write:


    If we plug in (4t) for x, we get

    So dy/dt=64t

    We know (dx/dt)=4 --> (dt/dx)=1/4 (since this denominator never goes to 0)

    so (dy/dx) = (dy/dt)*(dt/dx)=64t*(1/4) = 16t

    From the equation x=4t given above we know, t = (1/4)x

    so (dy/dx)= 16t= 16(x/4) = 4x

    Is this all right?


    Another confusing question for me:

    say we have x=3t

    Would d/dx(sinx)=cosx

    Or would it be


    When I see d/dx, I always think that just means, "take the derivative of this in terms of x only". What if we know x is a function of another variable though?
    Last edited: Sep 15, 2012
  2. jcsd
  3. Sep 15, 2012 #2


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    Yes, that is correct.

    Yes, that is still correct. We are differentiating with respect to x so any dependence on t is irrelevant.

    The middle term there looks suspicious. You should have (dx/dt)(d/dx)(2x2+3x)
    so it doesn't look like you want to differentiate (dx/dt) with respect to x. Other than that, yes, that is a correct application of the "chain rule". Notice that could also have replaced x with 2t:
    2x2+ 3x= 2(2t)2+ 3(2t)= 8t2+ 6t and the derivative of that is 16t+ t. Of course, that is the same as 2(4x+ 3)= 2(4(2t)+ 3)= 16t+ 6

    Yes, that is completely correct.
  4. Sep 15, 2012 #3
    Thanks again, hallsOfIvy. Since I have come to this forum you have answered so so many of my questions. I almost feel bad asking questions. I just hope you know how much I appreciate yours and others contributions. For myself I am just reviewing all of these concepts which are old to me, and it has helped my review so incredibly much to get feedback and insight from users here. I wish I could contribute more so I wouldn't just be taking taking taking and not giving, but unfortunately I am not yet confident enough in any answers to think I would be giving a correct answers to users who are submitting questions.
  5. Sep 15, 2012 #4
    Perhaps if I could ask one more question to make sure I've "got it"

    So, say we have a function F(t). We can say F(t) is an antiderivative of f(t) if d/dt(F(t))=f(t)

    Now if we have two such functions, does that mean for some u we can automatically say that:

    F(u) is an antiderivative of u, and that d/du(F(u))=f(u) ?

    For example,

    for f(x)=2x, we know that F(x)=x2 is an antiderivative, and that d/dx(F(x))=d/dx(x2)=2x=f(x)

    Now let's say, u=3t+4

    Does this mean that if we plug u in, and get

    f(u)=2(3t+4) and F(u)=(3t+4)2, that F(u) is still an antiderivative of f(u) and therefore


  6. Sep 15, 2012 #5
    The last statement [itex] = f(t) = f(u) [/itex] is incorrect, f(t) does not equal f(u).
    f(u) equals f(3t+4) which is not the same thing as f(t).

    Everything else looks good though.

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