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

Question

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 ?

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

Answer 1

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 $n$.

Answer 2

Lets derive it:

$\displaystyle f(x) = x e^{rx}$

$\displaystyle f'(x) = e^{rx}(rx + 1)$

$\displaystyle f''(x) = r e^{rx}(rx + 2)$

$\displaystyle f'''(x) = r^2e^{rx}(rx + 3)$

$\displaystyle \ldots$

$\displaystyle f^{(n)}(x) = r^{n-1}e^{rx}(rx + n) = \left(\frac{n}{r} + x \right)r^{n}e^{rx}$

Answer 3

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 $$ (fg)^{(n)}=\sum_{k=0}^{\color{red}1} {n\choose k} f^{(k)}g^{(n-k)} =f(x)g^{(n)}(x)+nf'(x)g^{(n-1)}(x)=xr^ne^{rx}+nr^{n-1}e^{rx}. $$