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Ichneumon
2009-10-21, 12:07 PM
So, I just have to check,

The derivative of f(x)=e^bx (with b being a constant number like -2, 5, 800 or something) would be b * e^bx, right?

Astrella
2009-10-21, 12:08 PM
Yep. (Assuming you're deriving to x of course.)

Cubey
2009-10-21, 12:10 PM
That's correct.

[g(f(x))]' = g'(f(x)) * f'(x)

chiasaur11
2009-10-21, 12:11 PM
As others have said:

Yes. Calc 1A?

Ichneumon
2009-10-21, 12:11 PM
Yep. (Assuming you're deriving to x of course.)

Thanks, had to be sure, my textbook ignores the examples with b in it, so I wanted to be sure that that happens to b. (Yes, of course, else everything would change accordingly). Thanks!



Yes. Calc 1A?

No, it's "Basic Math" course, needed for my study and e and logarithms confuse the hell out of me.

Mando Knight
2009-10-21, 12:26 PM
No, it's "Basic Math" course, needed for my study and e and logarithms confuse the hell out of me.

But exponential and sinusoidal functions are so nice when running differential equations!

Cobra_Ikari
2009-10-21, 12:45 PM
Related, and yet completely different question. Indefinite integral of e^x^n is done how? For example, e^x^2?

I remember one teacher telling me it couldn't be done, but I don't know if they meant "by what you will learn in this class", or period. Just curious.

RS14
2009-10-21, 01:06 PM
Related, and yet completely different question. Indefinite integral of e^x^n is done how? For example, e^x^2?

I remember one teacher telling me it couldn't be done, but I don't know if they meant "by what you will learn in this class", or period. Just curious.

Essentially as Cubey said. Let g(x)=e^x and f(x)=x^n. Then g(f(x))'=f'(x)*g'(f(x))=n(x^n-1)e^(x^n)

Cobra_Ikari
2009-10-21, 01:13 PM
Essentially as Cubey said. Let g(x)=e^x and f(x)=x^n. Then g(f(x))'=f'(x)*g'(f(x))=n(x^n-1)e^(x^n)

No, no. Not the derivative. Derivatives are easy. The indefinite integral. >.>

chiasaur11
2009-10-21, 01:17 PM
No, no. Not the derivative. Derivatives are easy. The indefinite integral. >.>

I think it's the "can't be done. Ever." one, from what I've heard in math related classes. Not sure, but...

douglas
2009-10-21, 01:26 PM
I specifically remember my high school calculus textbook stating that the indefinite integral of e^(-x^2) had been proven to be an impossible problem, so I'm pretty confident it's "can't be done, period." There might be some specific values of n that it's possible for (besides 1 and 0, of course), but the general problem is provably unsolvable.

Cobra_Ikari
2009-10-21, 01:29 PM
I specifically remember my high school calculus textbook stating that the indefinite integral of e^(-x^2) had been proven to be an impossible problem, so I'm pretty confident it's "can't be done, period." There might be some specific values of n that it's possible for (besides 1 and 0, of course), but the general problem is provably unsolvable.

Thank you. I was unclear whether unsolvable meant "with math you will learn in this class" or "for anyone".

You would assume it would have one, though. I guess it's just that we can't express it?

douglas
2009-10-21, 01:44 PM
It has an integral, certainly, every continuous function does, but it cannot be written out as a formula. We can only approximate definite integrals of it.

RS14
2009-10-21, 01:55 PM
No, no. Not the derivative. Derivatives are easy. The indefinite integral. >.>

Oh, that was stupid of me. Sorry.

In general, I believe you can give the integral in the form of an infinite power series (a_0+a_1x+a_2x^2...). That would be 1+x^2+x^4/2+... in this case. The integral of this is then x+x^3/3+x^5/10... (Note--I've not studied this since last spring, I may be mistaken).

It's difficult to work with, but is a solution.

You can also define functions that satisfy the integral, much as sine and cosine are defined. Properties can then be described. In this case, Mathematica reports that the integral of e^(x^2) is (((Pi)^(1/2))/2) *Erfi (http://en.wikipedia.org/wiki/Erfi)(x)

Tirian
2009-10-21, 02:16 PM
You would assume it would have one, though. I guess it's just that we can't express it?

It does have an indefinite integral; it's called the error function erf(x). (To be precise, erf(x) has a scalar multiplier so that erf(∞) = 1 instead of (√π)/2, which is more convenient for the folks who use it a lot.) There is a Taylor series, but no closed expression. Not really that much stranger than exp(x) or sin(x), I suppose.

douglas
2009-10-21, 02:19 PM
erf(x) doesn't really count because it's defined as the integral in question. Now if you find a way to express it that does not involve an integral or an infinite summation or product, that would be noteworthy.

Tirian
2009-10-21, 02:37 PM
*shrug* All I'm saying is that if you have no problem talking about ln(x) as being the indefinite integral of dx/x, then it's just the same thing to talk about erf(x) as being 2/√π times the indefinite integral of e^-x^2 dx. Both are defined as the definite integral of a formula.

douglas
2009-10-21, 03:18 PM
Er, no, ln(x) is defined as the inverse of e^x. It happens to also be the indefinite integral of dx/x, but that has nothing to do with its definition.

BugFix
2009-10-21, 05:20 PM
erf(x) doesn't really count because it's defined as the integral in question. Now if you find a way to express it that does not involve an integral or an infinite summation or product, that would be noteworthy.

As Tirian points out, that's not particularly notable. There's similarly no closed-form expression for calculating many other functions, like ln() or exp(), or sin(). Algebraic expressions are the exception, not the rule.