# Help! can someone differentiate e^(nx)

1. Nov 30, 2003

### Bailey

can someone differentiate e^(nx) where n is any integer. i think is equal to n*e^(nx).

please show the proof, thanx.

2. Dec 1, 2003

### sridhar_n

Well the proof is very big, so i just give u an idea as to how to go abt proving:

$$d(e^{(nx)})/dx = ne^{nx}$$

now increment y by a small value such that:

$$y + \Delta y = e^{(n(x + \Delta x))}$$

Divide the whole term by y such that:

$$(\Delta y)/y = [[e^{(n(x + \Delta x))}]/(e^{(nx)})] - 1$$ {We are putting $$d(e^{nx})/dx = n*e^{nx}$$ on the right hand side of the eqtn.}

take the one on the other side to get:

$$\Delta y/ y = [e^{(n(x + \Delta x))}- e^{(nx)}]/[e^{(nx)}]$$

now apply the limits $$\Delta y \longrightarrow 0$$ to bot sides and evaluate the limit to get your answer.

Sridhar

Last edited: Dec 1, 2003
3. Dec 1, 2003

### HallsofIvy

Staff Emeritus
Or, assuming you don't actually need to go back to the original definition of the derivative: apply the chain rule.

To differentiate y= enx, let u= nx.

Then y= eu so dy/du= eu

(If you don't know that then you will need to go back to definitions- in particular exactly how you are defining ex.)

du/dx= n, of course, and

dy/dx= (dy/du)(du/dx)= eu(n)= n enx.

4. Dec 1, 2003

### Bailey

thanx guys. i think the product rule can also use to differentiate
e^(nx), since e^(nx)=e^(n)*e^(x). but that will require much more time.

5. Dec 1, 2003

### lethe

this is not correct, the product rule does not apply here.

6. Dec 1, 2003

### lastlaugh

Hmm . .. I don't think that is quite legal since e^n is a constant, the product rule only applies if you have two functions.
you could alternately take the log of both sides which may look something like:

lny(x)=lne^nx
differentiate with respect to x
y'/y=n
y'=ny
and since y= e^nx
y'=n*e^nx

Of course I used the chain rule now on the left hand side, now

7. Dec 1, 2003

### futz

The product rule applies fine for a constant term, since a constant is a perfectly good function. It does not apply the way he said though; his exponential relation is wrong.

$$e^ne^x=e^{n+x}$$

8. Dec 1, 2003

### lastlaugh

It's mightily unnecessary as one term will automatically go to zero, and yes I missed his algebraic mistake

9. Dec 2, 2003

### Bailey

sorry about that mistake, what i meant was that,the product rule can be use repeatly to differentiate e^(nx).

eg. differentiate e^(3x)

e^(2x) = e^(x)*e^(x)
[e^(2x)]'=e^(x)*[e^(x)]' + e^(x)*[e^(x)]' = e^(x)*e^(x) + e^(x)*e^(x)
[e^(2x)]'= 2e^(2x)

e^(3x) = e^(2x)*e^(x)
[e^(3x)]'=e^(x)*[e^(2x)]' + e^(2x)*[e^(x)]' "sub [e^(2x)]'= 2e^(2x)"
[e^(3x)]'=e^(x)*(2e^(2x)) + e^(2x)*e^(x)
[e^(3x)]'=2e^(3x) + e^(3x)
[e^(3x)]'=3e^(3x)

10. Dec 2, 2003

### HallsofIvy

Staff Emeritus
Okay, you recognize that enx is NOT enex so that the a single application of the product rule does not work.

Yes, you can use the product rule repeatedly (in a proof by induction) but why?

What's wrong with using the chain rule as I suggested?

11. Dec 2, 2003

### HallsofIvy

Staff Emeritus
Actually, this is a more interesting problem that I first thought- there are several different ways of doing it.

The "obvious" way (to me anyway) is to note that the derivative of ex is ex and use the chain rule:
denx/dx= ex(d(nx)/dx)= nenx.

But we can also write enx= (en)x and use the fact that dax/dx= (ln a) ax:
d((en)x)/dx= ln(en)(en)x= nenx.

Or do it the other way around: enx= (ex)n) and use the power rule (together with the chain rule and the derivative of ex):
d((en)x)/dx= n((ex)n-1)(ex)= n(ex)n= nenx

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