I’m attempting to calculate d/dx e^x from first principles. My specific GQs at this point are a) can it be done and b) am I on the right track? I don’t want to Google it, because I don’t want the solution , and I’m hoping to avoid hints as well, other than the answers to a) and b).
I’ve gotten to:
(e^x) lim (Δx → 0) (1/Δx) (1/e^Δx -1).
If that doesn’t make sense, I’m happy to show my work if that would help.
So, the e^x is promising, but if I’m working it out correctly, the limit as Δx → 0 of 1/e^Δx is 1, so 1 - 1 = 0, which screws the whole thing up.
I’m assuming based on other such calculations I’ve seen (e.g., d/dx lnx from first principles) that there is some substitution or identity I have to do in order to make the limit work.
So, if the answer to a) and b) is yes, I’ll continue bashing my head against the wall. If a) is yes and b) is no, then I’ll start over. And if a) is no, then I’ll stop!
I’m rusty, but I think the “1/e^Δx” should be just “e^Δx”.
At which point you could do a power series expansion for e^x around zero, or be clever and think about what [e^(Δx-0) - e^0]/Δx is the definition of and substitute that in.
What first principles are you starting from? Because one of the more common definitions of e^x is that it’s the function of x which is its own derivative. And if you start from that principle, then of course the proof is trivial.
It all depends on how you define e^x. If you define it by the power series
e^x = 1 + x + x^2/2! + … + x^n/n! + …
then you can prove that this converges uniformly and thereby justify term by term differentation (not a triviality, but not that hard) then you will see that the derivative is the same power series. If you define e^x as the limit as n --> infinity of (1 + 1/n)^{nx}, then the problem is much harder and not easily answered here.
Slight hijack. Except for change of notation, Napier essentially defined natural logarithms as the solution to the differential equation df/dx = 1/x. From this it is straightforward to derive that the inverse function is its own derivative. But of course, calculus had not been sufficiently developed at that time to actually carry this out.
Damn, you’re right. And that’s what Quercus was getting at too.
Hari, so are you saying that I can’t get there from here, so to speak, assuming I’m trying to solve the problem as a limit as lim (Δx → 0) (f(x+Δx) - f(x))/Δx?
You can, but it depends on how you’re defining e^x.
And if you’re defining it in terms of exponentiation of some constant e (and there are multiple definitions for that constant, too) raised to some power, then you’re also going to need to define exponentiation to a non-integer power. The catch there is that the usual definition is in terms of e^x.
You’re going to need to use some definition of e at some point, and depending on which definition, it may be trivial.
But you can by first principles show the general form of derivatives of exponentials, though a real mathematician may point out a hidden tautology or multiple unstated assumptions here.
For easier notation let’s call f(x) = e^x and f’(x) = df(x)/dx. Then using the limit definition above you can pull e^x out of the limit, substitute in f(x) where appropriate and get
f’(x) = f(x) * lim (Δx → 0) (f(Δx-0) - f(0))/Δx. The part after the limit is just the definition of f’(0).
So f’(x) = f’(0)* f(x) This is true for any exponential n^x, not just e, of course. Again, one definition of e is that it’s the number for which f’(0) = 1.
No I think he wants to use just exponentials and normal algebra, and to get to THAT definition he’d first have to show it was a correct definition… which would involve the same question he has asked…
What I think he can do is realise that the log of the limit ought to be the limit of the log.
So do the limit of the log of that RHS that instead.
You can then use of log( a b ) = log a + log b and you can turn that ratio into a summation… and then its easy to do the limit …x
So now you have the limit of the log, the limit can be found.
e^x
That question from Ask Dr. Math seems odd, because my experience with calculus is that you start by defining the natural logarithm as a certain antiderivative of 1/x, e as the number whose natural log is 1, exp(x) the inverse of the natural logarithm, then noting e^x = exp(x) where e^x is defined, and thus consistently being able to state exp(x) = e^x for all x. This makes finding the derivative of e^x straightforward, although it takes a little work and the direct proof shown in that answer is just as short.
Thus I do not know in this alternate pedagogy how one defines e^x. Nor do I know the justification used for at least one step in that proof that involve passing an operation across the limits. Obviously, it’s justified because the function is well-behaved, but it sets a bad precedent if you’re going to be studying Calc at an advanced level into Real Analysis and beyond. (Of course, I’ve forgotten most of it since I left the academic math world, so maybe it’s simpler than I imagine, but any time I see an operation passing through a limit/integral, I’ve been trained to have to state why it’s justified, which a beginning calc class is not going to know anything about.)
The Calc textbook I had in college used this approach. It didn’t muck with the details in the main body of the text, but had appendices to cover the details. It was designed this way so the prof could either skip the details or cover them.
My prof covered them. It wasn’t bad at all in most respects. But it was a “gotcha” proof. It didn’t seem reasonable that anyone would have come up with this method early on but was worked out at some point by going backwards. I prefer methods based on the thought processes of someone trying to solve this for their first time. You can use modern concepts, but each step needs to be a plausible thing to try next.
BTW: It was made clear to me over and over that using power series/Taylor series type method at this level was not proper as those assumed you already knew the derivatives.
It’s just the first step in their proof: e = lim[n->inf] (1 + 1/n)^n
My personal experience is that e was defined that way before even getting to calculus because of its utility in the compound interest problem. For someone not yet used to derivatives, that definition still makes a lot of intuitive sense as it’s just what you’d expect from the question “if your investment has a 100% interest rate, what’s the return if the compounding is monthly, daily, hourly, etc.?”
Other definitions become nicer once you get to calculus, but the value of e is useful before that point.
A complete proof of this requires first defining the exponential function from first principles, which would first require defining the real numbers from first principles and each of these is a long hairy procedure. So any “proof” is something of a fake.
But assume all that is done and e is defined as lim_{h–>0}(1 + h)^{1/h} as in RadicalPi’s argument. He essentially reduced the problem, using the law of exponents, to showing that
lim_{h–>0}((e^h - 1)/h) = 1. So far so good. But in calculating e^h you cannot use the same variable as in the definition of e. Let me illustrate. lim_{h–>0}lim_{k–>0}k^h = lim_{h–>0}0=0, while, using l’Hospital rule, you can show that lim_{h–>0}h^h = 1.
So you would have to show that lim_{h–>0}lim_{k–>0}((1 + k)^{h/k} - 1)/h) = 1. Using some algebra this can be done for the case that h/k is a positive integer and argue using the monotonicity of the exponential function (which you had to have shown previously) that those cases suffice. A modification would work for negative exponents. But you can see that a full argument starting from first principles (the Peano postulates) would go on for page after boring page.
So I incline to defining the exponential by the power series, which you can easily show to be convergent for any x and do the algebra necessary to justify interchanging summation and differentiation. This renders the problem more or less trivial. Of course, the reals would have to be constructed and convergence of power series studied, but that is still far easier than what is outlined above.
When I posted the question, I indeed had something like what RadicalPi dug up in mind. And of course as is often the case when I ask these questions, I learned how much deeper the concepts are than I’m aware of!
What got me down this path is that Kahn academy has a video calculating the derivatives of ln x and e^x. For the derivative of ln x, he does the lim (Δx → 0) (f(x+Δx) - f(x))/Δx approach, but for e^x, he uses a very different method. So I just wondered why he did that and whether it was possible to do it in the way RadicalPi dug up.
Personally, I prefer to define exp() as the function which is its own derivative, and then show that the power series (which is a very convenient way of calculating it) matches that definition, then define ln() as the inverse function of exp(), then prove that integral(1/x) = log(x)+C. Then, you can prove that exp() matches the properties of exponentiation, exp(x) = e^x, where e is the constant defined as exp(1). And from there, you can define exponentiation of an arbitrary real number to an arbitrary real power in terms of exp().
But none of that is set in stone as the One True Way to do it. You can take any of those properties you like as definitional, and prove the rest from there.
You did get the sign wrong, and it doesn’t take you anywhere anyway, but note that the limit didn’t screw the whole thing up: the limit of Δx is also zero, so your limit point was d d/dx exp(x) = exp(x) 0 / 0; evaluation of the limit requires knowledge of the behaviour of exp(x) near zero.
e[sup]x[/sup] and ln x are the inverses of each other, so whichever derivative you prove first, you get the other one almost for free, via a form of the chain rule:
f**‘(y) = 1 / (f[sup]-1[/sup])’**(f(y))
h is the formal argument of lim (what is this type of variable called?) but is also being used outside the scope of the lim. I see that you get the right answer, but is there some comment needed to justify this step?
e^x = (1+x/n)^n (limit as n goes to infinity)
Chain rule and derivative of an expression raised to a power gives for derivative: n/n*(1+x/n)^(n-1) =
(1+x/n)^n / (1+x/n). As n goes to infinity, term in denominator goes to 1, resulting in e^x.
Anyway, I meant this one: