Derivative of the exponential function with base e

[Yh Hoo]

For an exponential function of the form y=a^x
First derivative , d/dx [a^x ] = a^x∙ lim┬(δx→0)⁡{(a^δx-a^0)/δx}
= a^x∙ m_((0,1))
now if m_((0,1)) which is the gradient of the y-axis intersection point of the exponential function equal to 1 exactly, then the derivative will be unchanged under differentiation and the value of the base a is then considered as e (Napier constant/natural number) .
IS there any derivation to get to the napier constant??

 

[jedishrfu]

check out wikipedia the e article shows many ways to determine its value. You would have to search a bit more to discover how these methods were derived.

http://en.wikipedia.org/wiki/E_(mathematical_constant)

 

[HallsofIvy]

If I understand your question, one way is this: for f(x)= a^x we have
\frac{f(x+h)- f(x)}{h}= \frac{a^{x+h}- a^x}{h}= \frac{a^xa^h- a^x}{h}= a<br /> ^x\frac{a^h- 1}{h}

Now, if we take the limit as h goes to 0, we have
\frac{da^x}{dx}= a^x\lim_{h\to 0}\frac{a^h- 1}{h}

So we will have the very nice property that de^x/dx= e^x if and only if
\lim_{h\to 0}\frac{e^h- 1}{h}= 1

Now, turn that around. I will just give a rough argument here- for h very close to 0, that limit says that (e^h- 1)/h= 1, approximately. Then e^h- 1= h, approximately. From that, e^h= 1+ h, approximately, so that e= (1+ h)^{1/h}. If we let n= 1/h, n goes to infinity as h goes to 0 and so the approximation e= (1+ 1/n)^n becomes exact as n goes to infinity:
e= \lim_{n\to\infty}\left(1+ \frac{1}{n}\right)^n
the "standard" limit definition of n.

Personally, I prefer to start by defining ln(x)= \int_1^x dt/t, then defining exp(x) to be the inverse function to ln(x), finally showing that exp(x)= exp(1)^x= e^x where e is now defined to be the value of x such that ln(x)= 1.

 

[Yh Hoo]

Thanks a lot! you really inspiring me!