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Math Help - I'm not getting e

  1. #1
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    I'm not getting e

    I finally figured out in the assignment I'm doing that one of the variables, known as e, is the e. I.e., it's not actually a variable.

    I gather (but can't prove) that of the properties of e is that"
    f(x) = e^x
    f'(x) = e^x

    Having forgotten that e is not just a regular variable, I decided to take the derivative the old-fashioned way:
    f(x) = e^x
    f'(x) = x \cdot e^{x-1}

    This leads me astray. Plugging in x=2, I find that ordinary derivation gives me:
    f(2) = e^2
    f'(2) = 2 \cdot e^{2-1} = 2e

    and I'm pretty sure

    e^2 = 2e
    only when e=2. And because this is the e, it doesn't equal 2!

    I can solve the problem now, only I have somewhat less faith in my new ability (?) to take a basic derivative. Is it just that ordinary derivation doesn't work with transcendental / irrational numbers, or did I derive wrong?
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  2. #2
    Bar0n janvdl's Avatar
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    Quote Originally Posted by Boris B View Post
    I finally figured out in the assignment I'm doing that one of the variables, known as e, is the e. I.e., it's not actually a variable.

    I gather (but can't prove) that of the properties of e is that"
    f(x) = e^x
    f'(x) = e^x

    Having forgotten that e is not just a regular variable, I decided to take the derivative the old-fashioned way:
    f(x) = e^x
    f'(x) = x \cdot e^{x-1}

    This leads me astray. Plugging in x=2, I find that ordinary derivation gives me:
    f(2) = e^2
    f'(2) = 2 \cdot e^{2-1} = 2e

    and I'm pretty sure

    e^2 = 2e
    only when e=2. And because this is the e, it doesn't equal 2!

    I can solve the problem now, only I have somewhat less faith in my new ability (?) to take a basic derivative. Is it just that ordinary derivation doesn't work with transcendental / irrational numbers, or did I derive wrong?

    Recall the chain rule.

    \frac{d}{du} e^{f(x)} = e^{f(x)} \cdot f'(x)
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  3. #3
    MHF Contributor
    Opalg's Avatar
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    Quote Originally Posted by Boris B View Post
    Having forgotten that e is not just a regular variable, I decided to take the derivative the old-fashioned way:
    f(x) = e^x
    f'(x) = x \cdot e^{x-1}

    This leads me astray.
    The reason it leads you astray is that the "old-fashioned way" doesn't apply here. It only tells you how to differentiate a function of the form x^a, where the base is the variable x, and the exponent is the fixed number a. But in the function e^x it's the other way round: the base is the fixed number e, and the exponent is the variable x. So you need a different technique altogether.

    Quote Originally Posted by Boris B View Post
    I gather (but can't prove) that of the properties of e is that
    f(x) = e^x
    f'(x) = e^x
    That's correct, and it's not surprising that you can't prove it, any more than you could prove the "old-fashioned" rule, unless a good teacher (or a good textbook) showed you how.
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