Why does

d/dx e^x = e^x

instead of

xe^(x-1)

This is a discussion on *d/dx e^x* within the **A Brief History of Cprogramming.com** forums, part of the Community Boards category; Why does
d/dx e^x = e^x
instead of
xe^(x-1)...

- 11-02-2003 #1
## d/dx e^x

Why does

d/dx e^x = e^x

instead of

xe^(x-1)

- 11-02-2003 #2
Because you can't use the power rule with functions raised to a variable.

d/dx k^x = k^(bx)*ln k*bAway.

- 11-02-2003 #3
you mean constants raised to a variable?

- 11-03-2003 #4
*Originally posted by DavidP*

**you mean constants raised to a variable?**

Instead of using a shortcut rule, actually put it through the long definition of a derivative and work it out.-Govtcheez

govtcheez03@hotmail.com

- 11-03-2003 #5
for me I did not understand it until I found a proof, so unless your going to look at a proof take it as is.

Man's mind once streched by a new idea, never regains its original dimensions

- Oliver Wendell Holmes

In other words, if you teach your cat to bark (output) and eat dog food (input) that doesn't make him a dog. It would have to chase cars, chew bones, and have puppies before I'd call it Rover ;-)

- WaltP

- 11-03-2003 #6

- Join Date
- Nov 2002
- Posts
- 1,109

just take it as it is. but if you want to know why, as cheez said, put it through the long definition of the derivative.

- 11-03-2003 #7
http://archives.math.utk.edu/visual....definition.12/ The definition of a derivative can be found there if you want to have a go at it

Away.

- 11-03-2003 #8

- Join Date
- Jan 2003
- Posts
- 1,708

so is this setup what cheez and you guys meant:

(e^(x+h) - e^(x)) / (h)

Lim h->0

I saw what confuted put for the answer but I'm not 100% sure the steps you have to take to get there.

EDIT: does b stand for base above? (change of base)Last edited by Silvercord; 11-03-2003 at 03:38 PM.

- 11-03-2003 #9
b and k were arbitrary constants.

(e^(x+h) - e^(x)) / (h)

Lim h->0

(e^((x+h)/x)) / (h)

Lim h->0

You'll have to take an ln() in there to get that to something you can work with.Away.

- 11-04-2003 #10
Use the Maclaurin series for e^x, and it becomes really obvious:

e^x = sum(k=0, infinity) [ x^k / k! ]

So for each term,

d[x^k / k!]/dx = k*x^(k-1) / k! = x^(k-1) / (k-1)

That is, when you take the derivative, each term becomes its predecessor, so the series remains unchanged.The word*rap*as it applies to music is the result of a peculiar phonological rule which has stripped the word of its initial voiceless velar stop.

- Exactly how to get started with C++ (or C) today
- C Tutorial
- C++ Tutorial
- 5 ways you can learn to program faster
- The 5 Most Common Problems New Programmers Face
- How to set up a compiler
- 8 Common programming Mistakes
- What is C++11?
- Creating a game, from start to finish

- How to create a shared library on Linux with GCC - December 30, 2011
- Enum classes and nullptr in C++11 - November 27, 2011
- Learn about The Hash Table - November 20, 2011
- Rvalue References and Move Semantics in C++11 - November 13, 2011
- C and C++ for Java Programmers - November 5, 2011
- A Gentle Introduction to C++ IO Streams - October 10, 2011