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[SOLVED] chain rule and e^(u)

DeusAbscondus

Active member
Jun 30, 2012
176
Hi folks,
I don't know if my experience is at all common (and I would like some feedback on this if possible), but I can't seem to nail down the properties of euler's number in the context of chain rule problems.

Here is the nub of my difficulty:

1. $\text{If }f(x)=e^x \text{then }f'(x)=e^x$

This I accept, though, not having seen a formal proof of it, and since it is counter-intuitive, I must take it on faith.
But the following I do not understand; could someone help me towards understanding?

2. $\text{If }g(x)=e^{x^{4}} \text{then }g'(x)=4e^{x^4}x^3$

Am I on the right track to observe that, if 2. is correct, then g(x) is a composite function, hence subject to the chain rule?

And if so, is the following a generally valid way to work through this and all such problems:

$\text{If }g(x)=e^{x^{4}} \text{find }g'(x)$

$\text{Now}g'(x)=u'v' \text{by Chain Rule}$

$\text{So, let }u=x^4 \text{and }v=e^u$

$\text{Then }u'=4x^3 \text{and }v'=e^u \text{ (by some rule which currently exceeds my understanding)}$

$\text{Therefore }g'(x)=u'v'=4x^3*e^u=\text{(via substitution) }4x^3*e^{x^{4}}$

$\text{Which, simplified }=4e^{x^{4}}x^3$


Finally, I have a similar hesitation/scruple/sense of vertigo when it comes to dealing with another unusual derivative, that of:

$ln(x)$

If anyone can see why, having the read the foregoing, I would feel unsure of myself around this animal, could they possibly add some notes to help me tame it?

Regs,
DeusAbs
 
Last edited:

Jameson

Administrator
Staff member
Jan 26, 2012
4,043
Hi DeusAbscondus, :)

1) Let's try to derive the derivative of $e^x$. Recall that for $f(x)$ it follows that \(\displaystyle f'(x) = \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\).

In this case that becomes \(\displaystyle \lim_{h \rightarrow 0} \frac{e^{x+h}-e^x}{h}=\frac{e^x \cdot e^h-e^x}{h}=\frac{e^x(e^h-1)}{h}\).

Ok, from here we should notice that $e^x$ isn't dependent on h in this limit thus can be brought outside of the computation as long as we multiply through at the end.

So the last bit becomes \(\displaystyle \left( e^x \cdot \lim_{h \rightarrow 0} \frac{e^h-1}{h} \right)\).

From here you need to compute that limit and you can do it a few ways. For a less rigorous approach that is more intuitive as well, I suggest graphing \(\displaystyle \frac{e^h-1}{h}\) on a graphing calculator and you'll see that as h approaches zero, then the y value approaches 1.

So at the end we get that \(\displaystyle \left( e^x \cdot \lim_{h \rightarrow 0} \frac{e^h-1}{h} \right) = e^x \cdot (1) = e^x\), therefore \(\displaystyle \frac{d}{dx} e^x = e^x\)

Please let me know if any of that is unclear. Again, it's not a rigorous proof by any means, more of a "walk through".

2) You are completely correct that \(\displaystyle \frac{d}{dx} e^{f(x)} = e^{f(x)} \cdot f'(x)\). As you said this is an application of the chain rule.
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
$\text{Then }u'=4x^3 \text{and }v'=e^u \text{ (by some rule which currently exceeds my understanding)}$
Hi DeusAbscondus, :)

The only rule that is used here is the chain rule.

\begin{eqnarray}

g(x)&=&e^{x^{4}}\\

g'(x)&=&\frac{d}{d(x^4)}e^{x^{4}}\frac{d}{dx}x^4\\

&=&e^{x^{4}}(4x^3)\\

\therefore g'(x)&=&4x^3e^{x^{4}}

\end{eqnarray}

Kind Regards,
Sudharaka.
 

Fantini

"Read Euler, read Euler." - Laplace
MHB Math Helper
Feb 29, 2012
342
Perhaps to avoid particularizing and help you think of the exponential as any other function, think of $e^{x^4}$ as $h(x) = e^x$ and $g(x) = x^4$, then $(h \circ g)(x) = h(g(x)) = e^{x^4}$ and $(h \circ g)' (x) = h'(g(x)) \cdot g'(x) = e^{x^4} \cdot (4x^3) = 4 e^{x^4} x^3.$
 

chisigma

Well-known member
Feb 13, 2012
1,704
Hi DeusAbscondus, :)

1) Let's try to derive the derivative of $e^x$. Recall that for $f(x)$ it follows that \(\displaystyle f'(x) = \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\).

In this case that becomes \(\displaystyle \lim_{h \rightarrow 0} \frac{e^{x+h}-e^x}{h}=\frac{e^x \cdot e^h-e^x}{h}=\frac{e^x(e^h-1)}{h}\).

Ok, from here we should notice that $e^x$ isn't dependent on h in this limit thus can be brought outside of the computation as long as we multiply through at the end.

So the last bit becomes \(\displaystyle \left( e^x \cdot \lim_{h \rightarrow 0} \frac{e^h-1}{h} \right)\).

From here you need to compute that limit and you can do it a few ways...
One possible 'rigorous' way to demonstrate that...

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

... is the following...

a) You start from the definition of exponential...

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

b) You demonstrate the identity...

$\displaystyle \lim_{n \rightarrow \infty} (1+\frac{h}{n})^{n}= \lim_{n \rightarrow \infty} \sum_{k=0}^{n} \frac {h^{k}}{k!} = 1 + h + \frac {h^{2}}{2}+... \frac{h^{n}}{n!}+...$ (3)

c) from (3) You derive...

$\displaystyle \frac{e^{h}-1}{h}= 1 + \frac{h}{2} +... + \frac{h^{n-1}}{n!}+... $ (4)

... and the (1) follows immediately...

Kind regards

$\chi$ $\sigma$
 

DeusAbscondus

Active member
Jun 30, 2012
176
thanks: Jameson, Sudharaka and Fantini: Jameson for the informal proof of $e^x$, Sudharaka for reminding me that there is no mystery here, just the chain rule and Fantini for the code behind $(h \circ g)$ and some other stuff besides.

Just a quick follow up:
can one of you gents refer me to a gentle approach to understanding a delta/epsilon proof, via video preferably, so as I can see it being worked out, pause it and study it?

and, Jameson, is this proof: the delta/epsilon proof, the more formal proof of the first derivative to which your notes allude?

thx again gentelmen,

DeusAbs
 

Jameson

Administrator
Staff member
Jan 26, 2012
4,043
It's not actually completely informal the way I did it, just not rigorous. By applying chisigma's technique as well as many other possibilities, such as L'Hopital's Rule you can soundly demonstrate that this limit is 1. A delta-epsilon proof of this limit wouldn't be asked in basic calculus course as far as I know.

Anyway, as to your question. I'm not sure about videos but here's a link that's part of a great Calculus help site I would recommend for anyone, Paul's Online Notes. The link I gave you specifically deals with delta-epsilon proofs and has nice diagrams plus examples that cover both the theory and show you how it's done in practice.

Let me know what you think! :)
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
thanks: Jameson, Sudharaka and Fantini: Jameson for the informal proof of $e^x$, Sudharaka for reminding me that there is no mystery here, just the chain rule and Fantini for the code behind $(h \circ g)$ and some other stuff besides.

Just a quick follow up:
can one of you gents refer me to a gentle approach to understanding a delta/epsilon proof, via video preferably, so as I can see it being worked out, pause it and study it?

and, Jameson, is this proof: the delta/epsilon proof, the more formal proof of the first derivative to which your notes allude?

thx again gentelmen,

DeusAbs
Hi DeusAbscondus, :)

A good place filled with video lessons is Khan Academy. Here is a link to the Epsilon Delta limit definition introduction video.

Kind Regards,
Sudharaka.
 

CaptainBlack

Well-known member
Jan 26, 2012
890
Hi folks,
I don't know if my experience is at all common (and I would like some feedback on this if possible), but I can't seem to nail down the properties of euler's number in the context of chain rule problems.

Here is the nub of my difficulty:

1. $\text{If }f(x)=e^x \text{then }f'(x)=e^x$

This I accept, though, not having seen a formal proof of it, and since it is counter-intuitive, I must take it on faith.
But the following I do not understand; could someone help me towards understanding?
From the way this is worded it seems that you think that there is some universally agreed definition of the exponential function. The truth is that there is not, you may have been told it is defined as:

\[\exp(x)=\lim_{n \to \infty} \left(1+\frac{x}{n}\right)^n\]

but it is often defined as the solution to the differential equation initial value problem: \(f'(x)=f(x), f(0)=1\), or as the inverse of the natural logarithm, ..

So to prove that the derivative of the exponential function is itself we really need to know what definition you are starting from (and also what you already know).

CB
 

DeusAbscondus

Active member
Jun 30, 2012
176

DeusAbscondus

Active member
Jun 30, 2012
176
From the way this is worded it seems that you think that there is some universally agreed definition of the exponential function. The truth is that there is not, you may have been told it is defined as:

\[\exp(x)=\lim_{n \to \infty} \left(1+\frac{x}{n}\right)^n\]

but it is often defined as the solution to the differential equation initial value problem: \(f'(x)=f(x), f(0)=1\), or as the inverse of the natural logarithm, ..

So to prove that the derivative of the exponential function is itself we really need to know what definition you are starting from (and also what you already know).

CB
I just had a look at graphs of these functions: given that they are mirror images of each other graphically, I would have thought their inter-definability (if i can put it thus) bordered the trivial.

I'd also have thought my post here (indeed, the combined oeuvre of my posts to date :cool:) makes perfectly clear my starting position in the knowledge of mathematics stakes: zero + epsilon.
Prescinding from that, any light you can throw on it would seem to me to be, well - in a dark place - illuminating, n'est-ce pas?

I always enjoy your sparse, ascetical take on things, Cap'n.

So, thanks for the input,
Deus Abs
 

DeusAbscondus

Active member
Jun 30, 2012
176
It's not actually completely informal the way I did it, just not rigorous. By applying chisigma's technique as well as many other possibilities, such as L'Hopital's Rule you can soundly demonstrate that this limit is 1. A delta-epsilon proof of this limit wouldn't be asked in basic calculus course as far as I know.

Anyway, as to your question. I'm not sure about videos but here's a link that's part of a great Calculus help site I would recommend for anyone, Paul's Online Notes. The link I gave you specifically deals with delta-epsilon proofs and has nice diagrams plus examples that cover both the theory and show you how it's done in practice.

Let me know what you think! :)
Thanks Jameson; just spent some hours there! I suffer from a dearth of really first-rate material and my course notes are a nest of errors and poorly edited material, so I appreciate such a rich fund of examples, proofs, worked examples and tips.

I'll be sure to make a fuller response once I've had more time to check it.

DeusAbs
 

chisigma

Well-known member
Feb 13, 2012
1,704
From the way this is worded it seems that you think that there is some universally agreed definition of the exponential function. The truth is that there is not, you may have been told it is defined as:

\[\exp(x)=\lim_{n \to \infty} \left(1+\frac{x}{n}\right)^n\]

but it is often defined as the solution to the differential equation initial value problem: \(f'(x)=f(x), f(0)=1\), or as the inverse of the natural logarithm, ..

So to prove that the derivative of the exponential function is itself we really need to know what definition you are starting from (and also what you already know).

CB
More that 280 years ago [!] Leonhard Euler defined in such way the exponential and natural log functions...

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

$\displaystyle \ln z= \lim_{n \rightarrow \infty} n\ (z^{\frac{1}{n}}-1)$ (1)

... and also demonstrated that the two function are inverse to one other. In my [humble ...] opinion the definitions (1) are also today 'the best' mainly because they are valid 'without additions' for any real or complex value of z...

Kind regards

$\chi$ $\sigma$
 

CaptainBlack

Well-known member
Jan 26, 2012
890
I'd also have thought my post here (indeed, the combined oeuvre of my posts to date :cool:) makes perfectly clear my starting position in the knowledge of mathematics stakes: zero + epsilon.
Prescinding from that, any light you can throw on it would seem to me to be, well - in a dark place - illuminating, n'est-ce pas?
You must have at least some idea of how the exponential function is defined, or at the very least how "e" is defined, otherwise you are starting in the middle.

(I would recommend you get a copy of Morris Kline's book "Calculus; An intuitive and physical approach", the only flaw of which is its use of US Customary Units)

CB
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,708
More than 280 years ago [!] Leonhard Euler defined in such way the exponential and natural log functions...

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

$\displaystyle \ln z= \lim_{n \rightarrow \infty} n\ (z^{\frac{1}{n}}-1)$ (1)

... and also demonstrated that the two function are inverse to one other. In my [humble ...] opinion the definitions (1) are also today 'the best' mainly because they are valid 'without additions' for any real or complex value of z...
In my [equally humble ...] opinion, the definitions (1) are not the best, because they are hard to work with. For example, starting from (1) how would you prove the index law $e^{x+y} = e^xe^y$, or the derivative rule $\frac d{dx}e^x = e^x$? Those proofs can surely be given, but the limit definition does not strike me as a natural or convenient one to start from. I much prefer the power series definition $e^x = \sum_{n=0}^\infty\frac{x^n}{n!}$ for the (complex) exponential. The basic properties follow quite easily from that, as does the fact that the real exponential $e^x:\mathbb{R}\to(0,\infty)$ increases monotonically and hence has an inverse, which defines the real logarithm function. (The complex logarithm is not a single-valued function, and must therefore be defined more indirectly.)
 

chisigma

Well-known member
Feb 13, 2012
1,704
In my [equally humble ...] opinion, the definitions (1) are not the best, because they are hard to work with. For example, starting from (1) how would you prove the index law $e^{x+y} = e^xe^y$, or the derivative rule $\frac d{dx}e^x = e^x$? Those proofs can surely be given, but the limit definition does not strike me as a natural or convenient one to start from. I much prefer the power series definition $e^x = \sum_{n=0}^\infty\frac{x^n}{n!}$ for the (complex) exponential. The basic properties follow quite easily from that, as does the fact that the real exponential $e^x:\mathbb{R}\to(0,\infty)$ increases monotonically and hence has an inverse, which defines the real logarithm function. (The complex logarithm is not a single-valued function, and must therefore be defined more indirectly.)
On the basis of the [easily enough demonstrable...] identity...

$\displaystyle \lim_{n \rightarrow \infty} (1+\frac{z}{n})^{n} = \lim_{n \rightarrow \infty} \sum_{k=0}^{n} \frac{z^{k}}{k!}$ (1)

... the two definitions must be considered fully equivalent. On the purely conceptual basis however the original Euler's definition requires only the concept of limit and is 'more elementary' than any other definition and that is an excellent reason for defining it as 'the best'... at least for me(Blush)..

Kind regards

$\chi$ $\sigma$
 
Jul 22, 2012
35
1. $\text{If }f(x)=e^x \text{then }f'(x)=e^x$
This is very easy to get using the power series definition from Opalg's post.

$\begin{aligned} & (e^x)' = \bigg(\sum_{k \ge 0}\frac{x^k}{k!}\bigg)' = \sum_{k \ge 0}\frac{kx^{k-1}}{k!} = \sum_{k \ge 1}\frac{kx^{k-1}}{k!} = \sum_{k+1 \ge 1}\frac{(k+1)x^{k}}{(k+1)!} = \sum_{k \ge 0}\frac{x^{k}}{k!}.\end{aligned}$
 

chisigma

Well-known member
Feb 13, 2012
1,704
This is very easy to get using the power series definition from Opalg's post.

$\begin{aligned} & (e^x)' = \bigg(\sum_{k \ge 0}\frac{x^k}{k!}\bigg)' = \sum_{k \ge 0}\frac{kx^{k-1}}{k!} = \sum_{k \ge 1}\frac{kx^{k-1}}{k!} = \sum_{k+1 \ge 1}\frac{(k+1)x^{k}}{(k+1)!} = \sum_{k \ge 0}\frac{x^{k}}{k!}.\end{aligned}$
That is true of course but [conceptually...] the preliminary demonstration of the convergence theorem of the series of derivatives is required. The computation of derivative of $e^{x}$ using the standard derivative definition require only to demonstrate that...

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

... and that requires elementary instruments...

Kind regards

$\chi$ $\sigma$