Thread: n-th derivative of u'exp(u)

Hello...

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

$\displaystyle
u'e^u\\
$

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

THANKS in advance

Originally Posted by paolopiace

Hello...

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

$\displaystyle
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...

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

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

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

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

I may be wrong, but I will assume this:

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

Originally Posted by colby2152

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

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

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

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

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

I may be wrong, but I will assume this:

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

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

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

Originally Posted by paolopiace

Hello...

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

$\displaystyle
u'e^u\\
$

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

THANKS in advance

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

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?

Originally Posted by paolopiace

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

$\displaystyle
u'e^u\\
$

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

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

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

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