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Derivative of e

  1. Nov 23, 2004 #1
    the derivative of e^u = e^u du/dx

    so the derivative of e would be 0 since:
    e^1 * 0 = 0

    is this correct?

    well, I knot it's not correct because derivative of e = e. It's because 1 is a constant right?

    derivative of e^(1+x) = e^(1+x) * 2

    is that correct?
    Last edited: Nov 23, 2004
  2. jcsd
  3. Nov 23, 2004 #2
    The derivative of e is 0 since e by itself is a constant (just like pi).

    The derivative of e^(1+x) is [e^(1+x)][1][dx].

    The derivative of e^x is (e^x)(dx)

    Hope that clarifies things up for you.
  4. Nov 23, 2004 #3
    "The derivative of e^(1+x) is [e^(1+x)][1][dx]."

    wouldn't it be The derivative of e^(1+x) is [e^(1+x)][1]+[dx].
  5. Nov 23, 2004 #4
    The derivative must always be taken with respect to some variable. You can't just say "take the derivative of e^u".

    Using u as the variable, the derivative df(u)/du of the function f(u) = e^u is e^u.
    Therefore the derivative of e^u with respect to u AT u = 1 is df(1)/du = e^1 = e.

    If u is actually a function itself, say u = g(x), then when we take the derivative of e^u, we are taking it with respect to x. Then e^u = e^(g(x)) = f(x)

    Now the derivative according to the chain rule is df(x)/dx = e^(g(x)).dg(x)/dx = e^u . du/dx. When you said that e^u du/dx = 0, this is only true if u is constant no matter what x is.

    You can see here that if f = e^u, df/du = e^u but df/dx = e^u du/dx, two different things.
  6. Nov 23, 2004 #5


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    Erk. Let's try to keep straight what we're doing here.

    The "derivative", all by itself, doesn't mean anything -- you need to say what you're differentiating with respect to.

    The derivative of the number "e" with respect to x is zero because e isn't a function of x (or anything else -- it's a constant).

    The derivative of the function ex with respect to x is the function ex, or more clearly,

    [tex]\frac{d(e^x)}{dx}|_u = e^u[/tex]

    In other words, the the derivative of ex with respect to x, evaluated at x=u, is the value eu.

    But please get those "dx" terms out of the derivative -- they have no business there. In basic differential calculus you should treat the "dx" as a reminder of what you're differentiating with respect to, and a nice reminder that it's a ratio of small values that you're talking about, but you shouldn't have it appearing by itself.

    Now as to that last derivative you asked about...

    [tex]\frac{d(e^{1+x})}{dx} = e^{1+x}[/tex]

    That's all -- no other terms appear, because

    [tex]\frac{d(1+x)}{dx} = 1[/tex]

    Now as to that "dx" -- you can call it a differential, in which case it's a linear function. You can call it a total derivative, but then you're talking about multivariable calculus, which isn't what you're asking about here. You can use it as the gradient 1-form of the basis function x, but that's really not what you're talking about here. Or you can use it as an "infinitesimal", which is the closest thing to what you're trying to do with it, but without a great deal more work to make "infinitesimals" into something rigorous it's just shorthand for the epsilon/delta limit process and you need to be careful to follow the rules if you use it that way. But if you want to end up with an actual function which can be evaluated at a particular point, which is what you're talking about when you say something like "the derivative of ex is ex", then you really need to keep the "dx" over there on the left with the derivative, and keep the result, over on the right, "clean".
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