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
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:
$\displaystyle f^{(n)}(u)=u^{(n-1)}e^u+(n+1)u^{(n)}u^{(n-1)}\cdots u^{(1)}e^u+\left(u^{(n+1)}\right)^{n-1}e^u$
There's a formula for taking the nth derivative of any function of the form y = f(x) g(x) called Leibniz's Rule.
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.