## Proofing the derivatives of e^x from the limit approach

I was searching for the proof of $$\frac{d}{dx} e^x = e^x$$.
and I found one in yahoo knowledge saying that

$$\frac{d}{dx} e^x = \lim_{Δx\to 0} \frac {e^x(e^{Δx}-1)} {Δx}$$

$$= \lim_{Δx\to 0} \frac {e^x [\lim_{n\to\infty} (1+ \frac{1}{n})^{n(Δx)}-1]} {Δx}$$

Let $$h= \frac {1}{n}$$, So that $$n = \frac {1}{h}$$

That makes
$$\lim_{Δx\to 0} \frac {e^x * [\lim_{h\to 0} (1+ h)^{\frac {1}{h}(Δx)}-1]} {Δx}$$

And it said variable h of the power $$\frac{1}{h}$$ and that of the limit $$\lim_{h\to 0}$$ of $$e^{Δx} = \lim_{h\to 0} (1+ h)^{\frac {1}{h}}$$ can be replaced by $$Δx$$, to make the whole formula becomes
$$= \lim_{Δx\to 0} \frac {e^x * [\lim_{Δx\to 0} (1+ Δx)^{\frac {1}{Δx} * {Δx}}-1]} {Δx}$$

because both variables are going to zero and that they can "replace" each other.

Of course, the calculation can still come to the answer that $$\frac{d}{dx} e^x = e^x$$.

But I wanna ask if it is true that, two different variables(neither of it is a function of another) approaching to the same value can be "replaced" by each other, or in other words, equal?
If that's true, what are the situations that this kind of "replacements" can't be done.
Also, if that's not true,is there any other way to prove that the derivatives of e^x =e^x without the use of any logarithms?

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 Not in general, I think. For example, $\lim \limits_{x \rightarrow 0} \frac { \lim \limits_{y \rightarrow 0} y } {x}$ is undefined. If you picked y = x, you'd get 1. You could, however, pick y = 2x as that too satisfies the arrow. I'm not sure why it was done here. Anyway, e^x can be derived if you just apply l'Hopital to the definition of derivative (your first line).
 Blog Entries: 27 Recognitions: Gold Member Homework Help Science Advisor hi ibelive! welcome to pf! (try using the X2 button just above the Reply box ) d/dx ( ∑ xn/n! ) = … ?

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## Proofing the derivatives of e^x from the limit approach

Yes, it's a tricky subject. I asked myself the same question on a different thread a few days ago. Anyway, you cannot use the series expansion (MacLaurin), unless you define e^x by the infinite sum.

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 Quote by dextercioby … unless you define e^x by the infinite sum.
i do!

 Recognitions: Science Advisor If you define the exponential as the inverse function of the integral of 1/x from 1 to x, everything follows immediately from the Chain Rule. the equation is x = $\int^{exp(x)}_{1}$1/xdx Differentiating wrsp to x shows that exp(x) is its own derivative. This though is using the logarithm without giving it a name. If you could show somehow that the derivative of exp is 1 at 0 then the rest follows instantly from the Newton quotient. i.e. exp'(x) = exp(x)lim h -> 0 (exp(h) - 1)/h So exp'(x) = exp(x)exp'(0)
 Blog Entries: 9 Recognitions: Homework Help Science Advisor This is interesting. But how would you prove it ?

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 Quote by hamsterman Not in general, I think. For example, $\lim \limits_{x \rightarrow 0} \frac { \lim \limits_{y \rightarrow 0} y } {x}$ is undefined. If you picked y = x, you'd get 1. You could, however, pick y = 2x as that too satisfies the arrow. I'm not sure why it was done here.
That's not quite correct. What you say is correct for
$$\lim_{(x,y)\to (0, 0)} \frac{y}{x}$$

But what you wrote,
$$\lim_{x\to 0}\frac{\lim_{y\to 0} y}{x}= 0$$
specifically takes the limit as y goes to 0 first, then takes the limit, as x goes to 0, of 0/x which is 0 for all non-zero x and so goes to 0.

 Anyway, e^x can be derived if you just apply l'Hopital to the definition of derivative (your first line).
But applying L'Hopital to the first line requires taking the derivative of $e^x$ doesn't it?

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 Quote by dextercioby This is interesting. But how would you prove it ?
Are you refering to Lavinia's post?

First define
$$ln(x)= \int_1^x \frac{1}{x} dx$$
Since 1/x is defined and continuous for all x except 0, it follows that this ln(x) exists, is continous and differentiable for all positive x. The derivative is, from the Fundamental Theorem of Calculus, 1/x. Since that derivative is positive, ln(x) is an increasing function.

By appropriate substitutions one can show that, for x and y both positive, ln(1/x)= -ln(x), ln(xy)= ln(x)+ ln(y), and, for x positive, $ln(x^y)= yln(x)$.

In particular, since ln(x) is differentiable for all positive x, it is, in particular, on the closed interval [1, 2] and we can apply the mean value theorem on that interval- there exist c in [1, 2] such that
$$\frac{ln(2)- ln(1)}{2- 1}= ln(2)= 1/c$$
And since $1\le c\le 2$, $1/2\le 1/c\le 1$. That is, $ln(x)\ge 1/2$. From that it follows that, for X any positive number, $ln(2^{2X})= (2X)ln(2)\ge (2X)(1/2)= X$. That is, ln(x) is unbounded above and, since it is increasing, $\lim_{x\to\infty} ln(x)= \infty$. From ln(1/x)= -ln(x), we have $\lim_{x\to 0} ln(x)= -\infty$.

That tells us that ln(x) maps the set of all positive numbers one-to-one and onto the set of all real numbers. That, in turn, means that it has an inverse function which we can call "exp(x)", a differentiable function from the set of all real numbers to the set of all positive numbers.

If y= exp(x), then x= ln(y) and so dx/dy= 1/y from which it follows that $dy/dx= y= e^x$.

Of course, I've just "called" that exp(x) and it is true that I have not shown that this is an "exponential". What we need is this:

Suppose y= exp(x). Then x= ln(y). If x is not 0, we can divide on both sides and have $1= (1/x)ln(y)= ln(y^{1/x})$. Going back to the exponential form, $y^{1/x}= exp(1)$ and so $y= exp(x)= (ln(1))^x$. If we define e= ln(1), we have $exp(x)= e^x$.

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 Quote by dextercioby This is interesting. But how would you prove it ?
Well all of the exponential functions f(x) = exp(ax) satisfy the relations

f(0) = 1 and f(x + y) =f(x)f(y).

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 Quote by lavinia Well all of the exponential functions f(x) = exp(ax) satisfy the relations f(0) = 1 and f(x + y+ =f(x)f(y). So you need more information to isolate exp(x) from the others.
Yes. You need to define e as f(1).

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 Quote by HallsofIvy Are you refering to Lavinia's post? First define $$ln(x)= \int_1^x \frac{1}{x} dx$$ Since 1/x is defined and continuous for all x except 0, it follows that this ln(x) exists, is continous and differentiable for all positive x. The derivative is, from the Fundamental Theorem of Calculus, 1/x. Since that derivative is positive, ln(x) is an increasing function. By appropriate substitutions one can show that, for x and y both positive, ln(1/x)= -ln(x), ln(xy)= ln(x)+ ln(y), and, for x positive, $ln(x^y)= yln(x)$. In particular, since ln(x) is differentiable for all positive x, it is, in particular, on the closed interval [1, 2] and we can apply the mean value theorem on that interval- there exist c in [1, 2] such that [tex]\frac{ln(2)- ln(1)}{2- 1}= ln(2)= 1/c. And since $1\le c\le 2$, $1/2\le 1/c\le 1$. That is, $ln(x)\ge 1/2$. From that it follows that, for X any positive number, $ln(2^{2X})= (2X)ln(2)\ge (2X)(1/2)= X$. That is, ln(x) is unbounded above and, since it is increasing, $\lim_{x\to\infty} ln(x)= \infty$. From ln(1/x)= -ln(x), we have $\lim_{x\to 0} ln(x)= -\infty$. That tells us that ln(x) maps the set of all positive numbers one-to-one and onto the set of all real numbers. That, in turn, means that it has an inverse function which we can call "exp(x)", a differentiable function from the set of all real numbers to the set of all positive numbers. If y= exp(x), then x= ln(y) and so dx/dy= 1/y from which it follows that $dy/dx= y= e^x$. Of course, I've just "called" that exp(x) and it is true that I have not shown that this is an "exponential". What we need is this: Suppose y= exp(x). Then x= ln(y). If x is not 0, we can divide on both sides and have $1= (1/x)ln(y)= ln(y^{1/x})$. Going back to the exponential form, $y^{1/x}= exp(1)$ and so $y= exp(x)= (ln(1))^x$. If we define e= ln(1), we have $exp(x)= e^x$.
This is all fine but more simply if you differentiate the equation

x = $\int^{exp(x)}_{1}$1/xdx

you get 1 on the left side and by the Chain Rule you get exp'(x)/exp(x) on the right.

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 Quote by lavinia This is all fine but more simply if you differentiate the equation x = $\int^{exp(x)}_{1}$1/xdx you get 1 on the left side and by the Chain Rule you get exp'(x)/exp(x) on the right.
Yes, but you still need to show that "exp(x)" really is a number to the x power.

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 Quote by HallsofIvy Yes. You need to define e as f(1).
I don't see this. Defining e as f(1) just gives different definitions of e as f varies. Unless you know e beforehand by some other definition. If you did then how would the proof go?

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 Quote by HallsofIvy Yes, but you still need to show that "exp(x)" really is a number to the x power.

Differentiating both sides gives you exp(x) = exp'(x)

I guess you are saying that you still need to show that exp(x + y) = exp(x)exp(y).

The same method works I think.

The derivative of f(x) = $\int^{exp(x)exp(y)}_{1}$1/xdx is again by the Chain Rule,

[1/exp(x)exp(y)]exp(x)exp(y) = 1 so f(x) = x + c since both functions have the same derivative. Setting x to zero shows that c = y.

But x + y = $\int^{exp(x + y)}_{1}$1/xdx

 Recognitions: Science Advisor The equation f'(x) = f'(0)f(x) shows that f is infinitely differentiable. The Taylor series looks like 1 + xf'(0) + .... 1/n!x$^{n}$f'(0)$^{n}$ + ..... This equation shows that each choice of f'(0) determines the function, f. Choosing f'(0) = 1 can be defined as the exponential function. One must then shows that the Taylor series converges for all x but this is not hard.
 Define: $\ln x \equiv \int_1^x \frac{1}{t} \, dt$ $\ln e\equiv 1$ By the fundamental theorem of calculus: $\frac{d}{dx}\ln x=\frac{1}{x}$ By the chain rule: $\frac{d}{dx}\ln x^n=\frac{1}{x^n}nx^{n-1}=\frac{n}{x}=\frac{d}{dx}(n\ln x)$ Two functions with equal derivatives over a domain differ by a constant over that domain: $\ln x^n=n\ln x+C \text{ for all } x>0$ Set $x=1$ $\ln 1^n=n\ln 1=0\Longrightarrow C=0$ So: $\ln x^n=n\ln x \text{ for all } x>0$ From above we have: $\ln e^x=x \ln e=x$ So: $\frac{d}{dx}\ln e^x=\frac{d}{dx}x=1=\frac{1}{e^x}\frac{d}{dx}e^x \Longrightarrow e^x=\frac{d}{dx}e^x$ q.e.d. Technically this only applies to rational numbers. We have to throw in some irrational spackle to make it apply to real numbers. I started out to prove this using limits, and realized that those proof are summed up in the fundamental theorem of calculus.