# Derivative of e^x from first principles

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• Jun 20th 2011, 05:22 AM
sakodo
Derivative of e^x from first principles
Hi guys,

Is it actually possible to derive the derivative of $\displaystyle e^x$ from first principles?

That is, to prove $\displaystyle \lim_{h \to 0}\frac{e^{x+h}-e^x}{h}=e^x$.

How do you go from there?

Or is it impossible because we actually define $\displaystyle e^x$ as a function whose derivative is $\displaystyle e^x$ itself?

Thanks.
• Jun 20th 2011, 05:28 AM
chisigma
Re: Derivative of e^x from first principles
Directly from the definition of exponential function You first demonstrate the 'fundamental limit'...

$\displaystyle \lim_{x \rightarrow 0} \frac{e^{x}-1}{x} = 1$ (1)

... and using (1) it is easy to demonstrate that is...

$\displaystyle \lim_{h \rightarrow 0} \frac{e^{x + h}- e^{x}}{h} = \lim_{h \rightarrow 0} e^{x}\ \frac{e^{h}-1}{h} = e^{x}$ (2)

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• Jun 20th 2011, 05:33 AM
Prove It
Re: Derivative of e^x from first principles
The function is DEFINED to be its own derivative. You CAN use this definition to evaluate the value of $\displaystyle \displaystyle e$ though.

\displaystyle \displaystyle \begin{align*} \lim_{h \to 0}\frac{e^{x + h} - e^x}{h} &= e^x \\ \lim_{h \to 0}\frac{e^xe^h - e^x}{h} &= e^x \\ \lim_{h \to 0}\frac{e^x(e^h - 1)}{h} &= e^x \\ e^x\lim_{h \to 0}\frac{e^h - 1}{h} &= e^x \\ \lim_{h \to 0}\frac{e^h - 1}{h} &= 1 \\ \lim_{h \to 0}(e^h - 1) &= \lim_{h \to 0}h \\ \lim_{h \to 0}e^h &= \lim_{h \to 0}(1 + h) \\ \lim_{h \to 0}(e^h)^{\frac{1}{h}} &= \lim_{h \to 0}(1 + h)^{\frac{1}{h}} \\ \lim_{h \to 0}(e) &= \lim_{h \to 0}(1 + h)^{\frac{1}{h}} \\ e &= \lim_{h \to 0}(1 + h)^{\frac{1}{h}} \end{align*}

And if you let $\displaystyle \displaystyle h = \frac{1}{n}$, then $\displaystyle \displaystyle n \to \infty$ as $\displaystyle \displaystyle h \to 0$, so we can write $\displaystyle \displaystyle e$ as its more common limit form

\displaystyle \displaystyle \begin{align*} e &= \lim_{h \to 0}(1 + h)^{\frac{1}{h}} \\ e &= \lim_{n \to \infty}\left(1 + \frac{1}{n}\right)^n \end{align*}
• Jun 20th 2011, 05:46 AM
chisigma
Re: Derivative of e^x from first principles
To demonstrate that...

$\displaystyle \lim_{x \rightarrow 0} \frac{e^{x}-1}{x}= 1$ (1)

... You start from the defintion of e...

$\displaystyle e= \lim_{\xi \rightarrow \infty} (1+\frac{1}{\xi})^{\xi}$ (2)

... and then set...

$\displaystyle e^{x} = 1+\frac{1}{\xi} \implies x= \ln (1+\frac{1}{\xi})$ (3)

... so that is...

$\displaystyle \lim_{x \rightarrow 0} \frac{e^{x}-1}{x} = \lim_{\xi \rightarrow \infty} \frac{\frac{1}{\xi}}{\ln (1+\frac{1}{\xi})}= \lim_{\xi \rightarrow \infty} \frac{1}{\ln (1+\frac{1}{\xi})^{\xi}}= \frac{1}{\ln e} = 1$ (4)

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• Jun 20th 2011, 05:51 AM
chisigma
Re: Derivative of e^x from first principles
An important detail: to define $\displaystyle e^{x}$ as the function that has itself as derivative is improper because any $\displaystyle f(x)=c\ e^{x}$ , where c is an 'arbitrary constant', has itself as derivative...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• Jun 20th 2011, 08:26 AM
CaptainBlack
Re: Derivative of e^x from first principles
Quote:

Originally Posted by chisigma
An important detail: to define $\displaystyle e^{x}$ as the function that has itself as derivative is improper because any $\displaystyle f(x)=c\ e^{x}$ , where c is an 'arbitrary constant', has itself as derivative...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

The definition you mean is: $\displaystyle e^x$ is the solution of the IVP

$\displaystyle f'(x)=f(x)$

with $\displaystyle f(0)=1$.

Which gets around the problem of the constant.

CB
• Jun 20th 2011, 11:17 AM
chisigma
Re: Derivative of e^x from first principles
May be that most of You don't agree with me but I'm strongly convinced that the only 'rigorous' mathematical arguments are based on the four elementary operations... so that, in my opinion of course, the only 'rigorous' definition of 'exponential function' is the definition given by Leonhard Euler two and half centuries ago...

$\displaystyle e^{x}= \lim_{n \rightarrow \infty} (1+\frac{x}{n})^{n}$ (1)

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• Jun 20th 2011, 01:26 PM
CaptainBlack
Re: Derivative of e^x from first principles
Quote:

Originally Posted by chisigma
May be that most of You don't agree with me but I'm strongly convinced that the only 'rigorous' mathematical arguments are based on the four elementary operations... so that, in my opinion of course, the only 'rigorous' definition of 'exponential function' is the definition given by Leonhard Euler two and half centuries ago...

$\displaystyle e^{x}= \lim_{n \rightarrow \infty} (1+\frac{x}{n})^{n}$ (1)

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

We may prefer one definition to another for aesthetic reasons but that does not make one better than the other in any absolute manner, nor does it make one more rigorous than the other.

I don't see that the limit definition is significantly closer to a definition based on "the four elementary operations... " than the differential equation definition or the inverse of the natural logarithm, etc...

CB
• Jun 20th 2011, 04:38 PM
HallsofIvy
Re: Derivative of e^x from first principles
Clearly, how you find the derivative of any function depends on how you define that function.

It is perfectly valid to define $\displaystyle ln(x)= \int_1^x \frac{1}{t}dt$. From that the fundamental theorem of calculus gives
$\displaystyle \frac{dln(x)}{dx}= \frac{1}{x}$.

We then define $\displaystyle e^x$ to be the inverse function to ln(x). If $\displaystyle y= e^x$ then $\displaystyle x= ln(y)$ so $\displaystyle \frac{dx}{dy}= \frac{1}{y}$ and then $\displaystyle \frac{dy}{dx}= y= e^x$
• Jun 20th 2011, 05:15 PM
chisigma
Re: Derivative of e^x from first principles
Quote:

Originally Posted by HallsofIvy
Clearly, how you find the derivative of any function depends on how you define that function...

It is perfectly valid to define $\displaystyle \ln x= \int_1^x \frac{1}{t}dt$...

All right!... in that case I propose to all You the following 'easy little task': demonstrate on the basis of the defintion above that is...

$\displaystyle \ln (x_{1}\ x_{2}) = \ln x_{1} + \ln x_{2}$ (1)

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• Jun 20th 2011, 09:01 PM
Deveno
Re: Derivative of e^x from first principles
i think it is perfectly "rigorous" to define $\displaystyle e^x$ in terms of power series (after all, we are dealing with real numbers, so convergence comes with the territory).

but this is the long way around, you have to define an infinite series, and what convergence for a series means, and then to show the usual power series for $\displaystyle e^x$ converges for all real x. then, you need to show that term-by-term differentiation is justified. having done all that, the proof of the derivative is easy.

the next best thing, is to define ln(x) as $\displaystyle \ln(x) = \int_1^x \frac{1}{t} dt$, and then define $\displaystyle e^x = \ln^{-1}(x)$. but then, to exhibit the derivative, one has to invoke the inverse function theorem (or dx/dy = 1/(dy/dx), which is what one does in practice, but is in my opinon, a little imprecise).

as for the proof of the so-called multiplicative property of ln(x) chisigma requested, here it is:

fix y > 0, and define for x > 0:

f(x) = ln(xy). then f'(x) = (ln'(xy))(xy)' (the chain rule)

= (1/(xy))(y) = 1/x. hence f' = ln', so f and ln differ by a constant, say c:

f(x) = ln(x) + c, for all x > 0. in particular, this holds for x = 1:

so ln(y) = ln(1y) = f(1) = ln(1) + c = c.

since y can be any positive real number,

ln(xy) = ln(x) + ln(y) for all x,y > 0.
• Jun 21st 2011, 01:15 AM
sakodo
Re: Derivative of e^x from first principles

Looking at posts #3 and #4 by Prove it and chisigma respectively, which one came first? The definition of $\displaystyle e=\lim_{n\to 0} (1+\frac{1}{n})^{n}$ or the defintion of $\displaystyle e^x$ as a function whose derivative is itself? Since if you use either one definition, you can always prove the other.

I think this is what caused my confusion.
• Jun 21st 2011, 03:20 AM
HallsofIvy
Re: Derivative of e^x from first principles
Very well, ChiSigma.

If we define $\displaystyle ln(y)= \int_1^x \frac{dt}{t}$, then it is immediate that ln(x) is defined, continuous, and differentiable for all x> 0. It is also immediate that $\displaystyle \frac{d ln(x)}{dx}= \frac{1}{x}$.

For x> 0, 1/x is also greater than 0 and we have $\displaystyle ln(1/x)= \int_1^{1/x}\frac{dt}{t}$. Make the substitution u= xt so that t= u/x and dt= du/x. Also, when t= 1, u= x, when t= 1/x, u= 1. Then $\displaystyle ln(1/x)= \int_x^1 \frac{x}{u}\frac{du}{x}= \int_x^1\frac{du}{u}= -\int_1^x \frac{du}{u}= -ln(x)$.

For x and y> 0, xy> 0 and we have $\displaystyle ln(xy)= \int_1^{xy}\frac{dt}{t}$. Make the substitution u= x/t so that t= xu and dt= xdu. Also, when x= 1, u= 1/x, when t= xy, u= y. Then $\displaystyle ln(xy}= \int_{1/x}^y \frac{1}{xu}xd= \int_{1/x}^y \frac{du}{u}= \int_{1/x}^1 \frac{du}{u}+ \int_1^y \frac{du}{u}$$\displaystyle = -\int_1^{1/x}\frac{du}{u}+ \int_1^y \frac{du}{u}= -ln(1/x)+ ln(y)= ln(x)+ ln(y)$.

For x> 0 and y any real number, $\displaystyle x^y$ is also positive and we have $\displaystyle ln(x^y)= \int_1^{x^y} \frac{dt}{t}$. If y is not 0, we can make the substitution $\displaystyle u= t^{1/y}$ so that $\displaystyle t= u^y$ and $\displaystyle du= yu^{y-1}du$. Also when t= 1, u= 1 and when $\displaystyle t= x^y$, $\displaystyle u= x$. $\displaystyle ln(x^y)= \int_1^x \frac{1}{u^y}(yu^{y-1}du= y\int_1^x \frac{du}{u}= yln(x)$.

If y= 0, then $\displaystyle x^y= x^0= 1$ so that $\displaystyle ln(x^y)= ln(1)= 0= 0(ln(x))= y ln(x)$ so $\displaystyle ln(x^y)= y ln(x)$ is true for any y.

Since $\displaystyle ln(x)$ is differentiable for all positive x, we can apply the mean value theorem on, say, the interval from x= 1 to x= 2. $\displaystyle \frac{ln(2)- ln(1)}{2- 1}= ln(2)= \frac{1}{c}$ where $\displaystyle 1\le c\le 2$. Then $\displaystyle \frac{1}{2}\le \frac{1}{c}\le 1$. That is, $\displaystyle ln(2)\ge \frac{1}{2}$.

That is important for the following reason: If X is any positive real number, then $\displaystyle ln(2^{2X})= 2X ln(2)\ge (2X)(1/2)= X$. That is, ln(x) is not bounded above. Since its derivative, 1/x, is always positive, ln(x) is an increasing function and since it is not bounded above, goes to infinity as x goes to infinity. From ln(1/x)= -ln(x), it is easy to see that ln(x) goes to negative infinity as x goes to 0. That shows that ln(x) maps the set of all positive real number "one to one and onto" the set of all real numbers and so has an inverse. We define exp(x) to be that inverse function.

One thing remains to be shown. If y= exp(x), then x= ln(y). If x is not 0, we can divide by it: $\displaystyle 1= \frac{1}{x}ln(y)= ln(y^{1/x})$. Going back to the exponential, $\displaystyle y^{1/x}= exp(1)$ so that $\displaystyle y= (exp(1))^x$. Of course, if x= 0, $\displaystyle y= exp(0)= 1= exp(1)^0$ so $\displaystyle exp(x)= (exp(1))^x$. That is, the "exp" function, defined as the inverse function to ln(x) really is some number to the x power. If we now define e to be exp(1) (the number whose natural logarithm is 1), then we have $\displaystyle exp(x)= e^x$.
• Jun 21st 2011, 03:23 AM
HallsofIvy
Re: Derivative of e^x from first principles
Quote:

Originally Posted by sakodo

Looking at posts #3 and #4 by Prove it and chisigma respectively, which one came first? The definition of $\displaystyle e=\lim_{n\to 0} (1+\frac{1}{n})^{n}$ or the defintion of $\displaystyle e^x$ as a function whose derivative is itself? Since if you use either one definition, you can always prove the other.

I think this is what caused my confusion.

"Which one came first" would be relevant to an historian but is not mathematically relevant. Any equivalent definition can be used.
• Jun 21st 2011, 04:25 AM
Plato
Re: Derivative of e^x from first principles
Quote:

Originally Posted by HallsofIvy
define $\displaystyle ln(x)= \int_1^x \frac{1}{t}dt$. From that the fundamental theorem of calculus gives
$\displaystyle \frac{dln(x)}{dx}= \frac{1}{x}$.

Quote:

Originally Posted by chisigma
All right!... in that case I propose to all You the following 'easy little task': demonstrate on the basis of the defintion above that is...
$\displaystyle \ln (x_{1}\ x_{2}) = \ln x_{1} + \ln x_{2}$

Is indeed an 'easy little task'.
Suppose that $\displaystyle a>0~\&~b>0$ then

$\displaystyle \ln (ab) = \int_1^{ab} {\frac{{dt}}{t}} = \int_b^{ab} {\frac{{dt}}{t}} + \int_1^b {\frac{{dt}}{t} = \underbrace {\int_1^a {\frac{{du}}{u}} }_{t = bu} + \int_1^b {\frac{{dt}}{t}} } = \ln (a) + \ln (b)$
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