Yea, I'm not sure what you are asking?
but
d(e) = 0
d(e^x) = e^x
d(e^-x) = -e^-x
2007-11-12 15:57:05
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answer #1
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answered by info2know 3
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I'm not sure I understand the question. e is a constant, and you took the derivative of a constant, you get zero.
But the exponential function, e^x is it's own derivative:
d/dx[e^x]=e^x.
It's fairly easy to prove using the definitions of limits. THere's also a proof using power series, but that comes a in a later course.
2007-11-13 00:05:33
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answer #2
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answered by thermodynaman2004 1
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If you're talking about the DERIVATIVE of e^x (with respect to x) then it (the derivative) is itself (the very function that you're taking the derivative of): e^x.
If you're talking about some constant e (typically e = 2.7183 in math) then the derivative of a constant with respect to an independent variable is zero.
2007-11-13 00:02:09
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answer #3
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answered by answerING 6
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e is a constant: 2.718281828....
Like every other constant, it's derivative is 0. However the derivative of e^x is e^x. Is that what you are asking?
2007-11-12 23:58:39
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answer #4
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answered by John B 6
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e
2007-11-12 23:57:30
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answer #5
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answered by Harris 6
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e is its own differential
f'(e) = e
the rule for e is....
e^ax, where a is a constant, and x is the variable
f'(x) = a(e^ax)
f''(x) = a[f'(x)] = a[a(e^ax)] = (a^2)(e^ax)
2007-11-12 23:56:17
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answer #6
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answered by jacobrcotton 3
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e could be anything. Please make your question more concrete.
2007-11-12 23:55:43
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answer #7
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answered by Jun Agruda 7
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