# $\frac{ \mathrm d^n xe^ {rx}}{\mathrm dx^n} \stackrel{?}{=} ( \frac {n}{r} + x )(r^n)e^{rx}$

I just found this equality on the internet, and I am wondering is there a way to prove it or where I can find the proof ?

$$\displaystyle \frac{ \mathrm d^n}{\mathrm dx^n} {xe^ {rx}} \stackrel{?}{=} ( \frac {n}{r} + x )(r^n)e^{rx}$$

-
See section $6$ in this paper for a general technique and examples. – Mhenni Benghorbal Mar 9 '13 at 16:25
By the way, you can try MAPLE to find the nth derivative of some functions with the command diff( f(x), x$n ). – Mhenni Benghorbal Mar 9 '13 at 16:30 Related problem 1, problem 2. – Mhenni Benghorbal Mar 9 '13 at 16:55 ## 3 Answers You may use Leibniz's rule $$(fg)^{(n)}=\sum_{k=0}^n {n\choose k} f^{(k)}g^{(n-k)},$$ where$f^{(k)}$means the$k$-th derivative of a function$f$(by convention,$f^{(0)}=f$). Put$f(x)=x$and$g(x)=e^{rx}$. Then$f^{(k)}(x)=0$for$k>1$. Hence Leibniz's rule gives $$(fg)^{(n)}=\sum_{k=0}^{\color{red}1} {n\choose k} f^{(k)}g^{(n-k)} =f(x)g^{(n)}(x)+nf'(x)g^{(n-1)}(x)=xr^ne^{rx}+nr^{n-1}e^{rx}.$$ - Suggestion: How about a proof by induction on$n$? 1. Prove the equality holds for the base case$n = 1$2. Assume the equality holds for$n = k, \;\;$(The Inductive hypothesis) 3. Show, using the inductive hypothesis, that the equality holds for$n = k+1$. That taking the derivative of the of the the$k^{th}$derivative gives you the equality of the form needed for$n = k+1$. Then, once you show you've shown your base case to hold$(1)$, and once you've completed$(3)$, you are justified in concluding that the equation holds for all$n$. - Let me know if you're familiar with induction proofs. – amWhy Mar 9 '13 at 4:18 Lets derive it:$\displaystyle f(x) = x e^{rx}\displaystyle f'(x) = e^{rx}(rx + 1)\displaystyle f''(x) = r e^{rx}(rx + 2)\displaystyle f'''(x) = r^2e^{rx}(rx + 3)\displaystyle \ldots\displaystyle f^{(n)}(x) = r^{n-1}e^{rx}(rx + n) = \left(\frac{n}{r} + x \right)r^{n}e^{rx}\$

-
Excellent...show the pattern...generalize! +1 – amWhy Apr 23 '13 at 0:38
@amWhy: While I am at it, I will the same about your answer +1 :-) – Amzoti Apr 23 '13 at 0:46