Derivative of e^x = e^x

  • Thread starter MHD93
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  • #1
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Hello
As I was trying to figure out why the derivative of e^x = e^x using the derivative definition I faced this limit:

lim (e^x - 1)/x ; x goes to 0

and I need your help, thank you.
 

Answers and Replies

  • #2
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Have you tried applying L'Hopitals rule?

That should clear things up.
 
  • #3
hunt_mat
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Have you tried the power series for exp(x)?
[tex]
e^{x}=1+x+\frac{x^{2}}{2}+\cdots
[/tex]
 
  • #4
hunt_mat
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jegues, in order to do that you would have to know the derivative of exp(x) which is what he was trying to calculate in the first place.
 
  • #5
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#Jegues: I don't know L'Hopitals rule, but thanks for reply

#hunt mat: thank you so much, this made the problem so easy
 
  • #6
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Mohammad_93, one of the ways the exponential function is defined is by the fact that its derivative of this function is the exponential function itself !
 
  • #7
HallsofIvy
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For any positive real number, a, the derivative of [itex]a^x[/itex] is given by:

[tex]\lim_{h\to 0}\frac{a^{x+h}- a^x}{h}= \lim_{h\to 0}\frac{a^xa^h- a^x}{h}[/tex]
[tex]= \lim_{h\to 0}a^x\frac{a^h - 1}{h}= a^x\left(\lim_{h\to 0}\frac{a^h- 1}{h}[/tex]

That is, the derivative of [itex]a^x[/itex] is [itex]a^x[/itex] itself times a constant, [itex]C_a[/itex]. (Constant in the sense that it does not depend on x. It does, of course, depend on a.)

It is easy to show that, for a= 2 or a= 2.5, say, that constant is less than 1. it is equally easy to show that for a= 3 or 4, say, the constant is larger than 1. We define "e" to be the number such that
[tex]\lim_{h\to 0}\frac{e^h- 1}{h}= 1[/tex]

Another, in my opinion, simpler way to show that the derivative of [itex]e^x[/itex] is [itex]e^x[/itex] is to first define
[tex]ln(x)= \int_1^x \frac{1}{t}dt[/tex]

It is easy to see that the derivative of ln(x), by the fundamental theorem of calculus, is 1/x. You can also show that it is a 1 to 1 function from the set of positive real numbers to the set of all real numbers and so has an inverse function, exp(x), from the set of real numbers to the set of positive real numbers. One can then show that exp(x) has derivative exp(x) and that [itex]exp(x)= (exp(1))^x[/itex] where exp(1) is defined, of course, by ln(exp(1))= 1.
 
  • #8
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Let f(x)=e^x

Clearly:
(e^x - 1)/x = (e^x - 1)/(x-0) = ( f(x) - f(0) )/x-0 ----> f'(0) as x--> 0
 
  • #9
HallsofIvy
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Let f(x)=e^x

Clearly:
(e^x - 1)/x = (e^x - 1)/(x-0) = ( f(x) - f(0) )/x-0 ----> f'(0) as x--> 0
Yes, that's true but not relevant to the question of what that derivative is. The last part, (f(x)- f(0))/(x- 0)----> f'(0), is true for any function that is differentiable at x= 0. The question, really, was "How do you show that [itex]lim_{x\to 0} (e^x-1)/x= 1[/itex]".
 
  • #10
Char. Limit
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Another, in my opinion, simpler way to show that the derivative of [itex]e^x[/itex] is [itex]e^x[/itex] is to first define
[tex]ln(x)= \int_1^x \frac{1}{t}dt[/tex]

It is easy to see that the derivative of ln(x), by the fundamental theorem of calculus, is 1/x. You can also show that it is a 1 to 1 function from the set of positive real numbers to the set of all real numbers and so has an inverse function, exp(x), from the set of real numbers to the set of positive real numbers. One can then show that exp(x) has derivative exp(x) and that [itex]exp(x)= (exp(1))^x[/itex] where exp(1) is defined, of course, by ln(exp(1))= 1.

You don't even need to define log(x) that way. It's easy to prove that using the basic definition of e, as follows:

[tex]e = lim_{x\rightarrow\infty}\left(1+\frac{1}{x}\right)^x[/tex]

And using the definition of the natural log, again as follows:

[tex]IF x=e^y\\ THEN y = log(x)[/tex]

Then by applying the definition of a derivative, you can prove, algebraically, that the derivative of log(x) is 1/x. Then you can use that to prove the derivative of e^x without ever using an integral or the FTC.
 

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