Proof that the derivative of e^x is itself

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The derivative of e^x is itself but have you ever wondered why this is true?

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I remember in school E was introduced definitially as the base for which e to the power x is it's own derivative, followed by being asked to use a calculator to try and approximate e, after tehy showed a few bases and how derivatives shift, which clearly implied there is some value for which this held

shadeblackwolf
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Circular logic in lemma 2 where you take d/dx e^y = e^y dy/dx

chillfill
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Hey pal, here from your dad's channel, great channel btw I need help in math so I'm here to stay.

joshuatony.
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*@ Noom Mood* --- Please write on paper *without* grid lines, because the grid lines make it
more difficult to read.

robertveith
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You can use the limit definition of e^x and just take the derivative and see it is itself

dqrksun
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Hey bro! Are you learning maths just for fun, or do you have some aim in mind, like Mathematical Olympiad? And how do you learn those things? Books, internet maybe?

kamiljan
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why did you add n in its exponent to get its proof?

kyllechristophermendiola
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Well, exp(x) is defined as a fonction being its own derivative and exp(0)=1, so there is no proof needed, especially because all of exp's properties comme from this definition

paulchiche
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This was very hard to follow what you were trying to do

JackTheAwesomeKnot