# Induction to prove a simple formula for calculating the $n$-th derivative?

I have posted about induction on it's own so I think I understand that part. But I'm slightly confused when it comes to using induction to show that $n+1$ is also the derivative. The base case if fine, I just set $n=1$. And show that is equal to the derivative not using the formula. But how about $k+1$? Can I just choose $k=$the-basecase. And than prove $k+1$ by just finding $f''(x)$?

$f^{(n)}(x) = a^{n-1}e^{ax}(ax+n)$

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We have $f^n(x)= a^{n-1}e^{ax}(ax+n)$ and so

$$\frac{d}{dx} f^n(x) = a^{n-1} \frac{d}{dx} \lbrace e^{ax}(ax+n) \rbrace = a^{n-1} \left( ae^{ax}(ax+n) + ae^{ax} \right)$$

$$= a^n e^{ax}(ax + n+1) = f^{n+1}(n).$$

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LEMMA $\rm\quad\quad\ \ (f_0\:e^{\:g})^{(n)}\ \ \ \ =\ \ f_n\ e^g\ \:$ when $\rm\: \ g'\ f_n + f_n'\: =\ f_{n+1}$

Base case: $\rm\quad\ (f_0\:e^{\:g})^{(0)}\ \ \ \ =\ \ f_0\ e^g\ \$ is true.

Induction: $\rm\quad (f_0\:e^{\:g})^{(n+1)} = \ (f_n\ e^g)'\ \:=\ (g'\ f_n + f_n')\ e^g \ =\ f_{n+1}\ e^g\quad\$ QED

Now $\rm\ g = ax,\ f_{n+1} = a^n\:(ax+n+1)\ =\ a\ (a^{n-1} (ax + n))+ a^n\ =\ g'\ f_n + f_n'$

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