So as stated in the title I have to create a general formula for nth derivative of $f(x) = \cos(ax)$ and proving it with induction.
I get $f(x) = \cos(ax)$
$f'(x) = -a\sin(ax)$
$f''(x) = -a^2\cos(ax)$
$f'''(x) = a^3\sin(ax)$
$f''''(x) = a^4\cos(ax)$
How are you supposed to algrebraically write a formula for this? I obviously see a pattern but I don't know how to approach this. Any help would be greatly appriciated.
Best Answer
When $n=1$, you get $-a \sin ax$.
When $n=3$, you get $a^3 \sin ax$.
When $n=5$, you get $-a^5 \sin ax$.
So at each step you multiply by $-a^2$. So $f^{(n)} (x)=(-a^2)^{(n-1)/2}(-a)\sin ax,$ for $n$ odd.
When $n=0$, you get $ \cos ax$.
When $n=2$, you get $-a^2 \cos ax$.
When $n=4$, you get $a^4 \cos ax$.
So at each step you multiply by $-a^2$. So $f^{(n)} (x)=(-a^2)^{n/2}\cos ax,$ for $n$ even.