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Thread: Derivata

  1. #1 Derivata 
    Forum Masters Degree thyristor's Avatar
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    If f(x)=e<sup>x</sup> why is f' equal to e<sup>x</sup>?


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  3. #2  
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    When you take the derivative of a function and evaluate it at a point, you obtain the rate of change, or slope, at that point. The graph of e<sup>x</sup> has a slope of one at 0. Thus, the derivative of e<sup>x</sup> is e<sup>x</sup> because if you evaluate it at 0, you obtain e<sup>0</sup> = 1. I've read that the e<sup>x</sup> graph is actually defined by having a slope of 1 at x = 0. I'm sure someone can give a better answer, but that's how I know it.


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  4. #3 Re: Derivata 
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    Quote Originally Posted by thyristor
    If f(x)=e<sup>x</sup> why is f' equal to e<sup>x</sup>?
    Because

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  5. #4  
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    Also note that, in Jane's description, the factor e<sup>x</sup> pulls out of the limit, so that you can conclude that the derivative of e<sup>x</sup> is:

    e<sup>x</sup> lim<sub>h->0</sub> (e<sup>h</sup>-1)/h

    So then the fact that e<sup>x</sup> is its own derivative boils down to the fact that the above limit is 1.
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  6. #5 Re: Derivata 
    Forum Masters Degree thyristor's Avatar
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    Quote Originally Posted by JaneBennet
    Quote Originally Posted by thyristor
    If f(x)=e<sup>x</sup> why is f' equal to e<sup>x</sup>?
    Because

    I don't understand how you can see that that as an answer to my question. When I asked my question I wanted to be answered something containing the very special definition of e. The thing I'm curious about is why not all functions a<sup>b</sup> have the derivata a<sup>b</sup>.
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  7. #6  
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    You want to do it another way? Try



    Now differentiate term by term. What do you get?
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  8. #7  
    Forum Professor river_rat's Avatar
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    You can also look at it from the opposite side thyristor, think of the logarithm ln(x) = ∫ dt/t where we integrate from 1 to x. Now a quick look at the chain rule shows that (e<sup>x</sup>)' = e<sup>x</sup>.

    However, if you think of what happens to a logarithm when you change the basis of that logarithm you will see that you introduce a constant that you just can't get rid of. This constant is what ruins this property for the other exponents. In fact, e is that number such that ∫ dt/t where we integrate from 1 to e is equal to 1 (so the constant is no problem )
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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  9. #8  
    Forum Masters Degree thyristor's Avatar
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    Quote Originally Posted by JaneBennet
    You want to do it another way? Try



    Now differentiate term. What do you get?
    I see, it's the same
    1+x+x<sup>2</sup>/2!+x<sup>3</sup>/3!...
    Quite neat.
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  10. #9  
    Forum Masters Degree thyristor's Avatar
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    Thanks for the help everybody :-D
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