n-th derivative of u'exp(u)

**paolopiace** · Jan 31st 2008, 10:41 AM

Hello...

BIG question (big for me, at least): Is there a general formula for the n-th derivative of

$ u'e^u\\ $

Basically I have a function exactly like that and I need to extract the n-th derivative.

THANKS in advance

**colby2152** · Jan 31st 2008, 11:07 AM

Originally Posted by paolopiace

Hello...

BIG question (big for me, at least): Is there a general formula for the n-th derivative of

$ u'e^u\\ $

Basically I have a function exactly like that and I need to extract the n-th derivative.

THANKS in advance

I noticed a pattern at the 2nd derivative...

$f(u)=u'e^u \Rightarrow u^{(1)}e^u$

$f^{(1)}(u)=u^{(2)}e^u+(u^{(1)})^2e^u$

$f^{(2)}(u)=u^{(3)}e^u+u^{(2)}u^{(1)}e^u + 2u^{(2)}u^{(1)}e^u+\left(u^{(3)}\right)^2e^u$

$f^{(2)}(u)=u^{(3)}e^u+3u^{(2)}u^{(1)}e^u+\left(u^{ (3)}\right)^2e^u$

I may be wrong, but I will assume this:

$f^{(n)}(u)=u^{(n-1)}e^u+(n+1)u^{(n)}u^{(n-1)}\cdots u^{(1)}e^u+\left(u^{(n+1)}\right)^{n-1}e^u$

**mr fantastic** · Jan 31st 2008, 09:27 PM

Originally Posted by colby2152

I noticed a pattern at the 2nd derivative...

$f(u)=u'e^u \Rightarrow u^{(1)}e^u$

$f^{(1)}(u)=u^{(2)}e^u+(u^{(1)})^2e^u$

$f^{(2)}(u)=u^{(3)}e^u+u^{(2)}u^{(1)}e^u + 2u^{(2)}u^{(1)}e^u+\left(u^{(3)}\right)^2e^u$

$f^{(2)}(u)=u^{(3)}e^u+3u^{(2)}u^{(1)}e^u+\left(u^{ (3)}\right)^2e^u$

I may be wrong, but I will assume this:

$f^{(n)}(u)=u^{(n-1)}e^u+(n+1)u^{(n)}u^{(n-1)}\cdots u^{(1)}e^u+\left(u^{(n+1)}\right)^{n-1}e^u$

There's a formula for taking the nth derivative of any function of the form y = f(x) g(x) called Leibniz's Rule.

**paolopiace** · Jan 31st 2008, 10:35 PM

...to see if there is a more simple and straightforward rule for u'exp(u).

**curvature** · Jan 31st 2008, 10:42 PM

Originally Posted by paolopiace

Hello...

BIG question (big for me, at least): Is there a general formula for the n-th derivative of

$ u'e^u\\ $

Basically I have a function exactly like that and I need to extract the n-th derivative.

THANKS in advance

Since $u'e^u$ is the derivative of $e^u$ , you only need to consider the $(n-1)$ -th derivative of $e^u$ .

**colby2152** · Feb 1st 2008, 05:04 AM

Originally Posted by mr fantastic

There's a formula for taking the nth derivative of any function of the form y = f(x) g(x) called Leibniz's Rule.

Yeah, I never really used that rule. It is handy though - it is basically a binomial expansion, but for derivatives. Did I get the derivative right, or are there more terms in the middle?

**Opalg** · Feb 1st 2008, 06:17 AM

Originally Posted by paolopiace

Is there a general formula for the n-th derivative of

$ u'e^u\\ $

Basically I have a function exactly like that and I need to extract the n-th derivative.

If $y = u'e^u$ then $y^{(n)} = D_n(u)e^u$ , where $D_n(u)$ is some polynomial in u and its derivatives. If you differentiate one more time then you get $y^{(n+1)} = (D_n'(u) + u'D_n(u))e^u$ . So $D_{n+1} = D_n'(u) + u'D_n(u)$ . If you apply this recursive formula to find the first few derivatives then you get

$y^{(1)} = y' = \bigl((u^{(1)})^2 + u^{(2)}\bigr)e^u$ ,
$y^{(2)} = \bigl((u^{(1)})^3 + 3u^{(1)}u^{(2)}+ u^{(3)}\bigr)e^u$ ,
$y^{(3)} = \bigl((u^{(1)})^4 + 4u^{(1)}u^{(3)}+ 6(u^{(1)})^2u^{(2)} + 3(u^{(2)})^2 + u^{(4)}\bigr)e^u$ ,
$y^{(4)} = \bigl((u^{(1)})^5 + 10(u^{(1)})^3u^{(2)} + 14(u^{(1)})^2u^{(3)} + 15u^{(1)}(u^{(2)})^2 + 5u^{(1)}u^{(4)} + 6u^{(2)}u^{(3)} + u^{(5)}\bigr)e^u$ .

That should be enough to convince you that there is not going to be any simple formula for the n-th derivative.

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