By $n^{th}$ order product rule, we find

$$f^{(50)}=\sum_{k=0}^{50}(x^2)^{(k)}\cos^{(50-k)}(ax)$$

Because $(x^2)' = 2x$, $(x^2)''=2$ and $(x^2)'''=0$, the above equation has at most $3$ terms.

We also note that the derivatives of $\cos{x}$ alternate among $\pm\sin{x}$ and $\pm\cos{x}$:

• $(\cos{x})^{(1)}=-\sin{x}$
• $(\cos{x})^{(2)}=-\cos{x}$
• $(\cos{x})^{(3)}=\sin{x}$
• $(\cos{x})^{(4)}=\cos{x}$

It follows that

• $(\cos{x})^{(50)}=-\cos{x}\implies (\cos(ax)^{(50)}=-a^{50}\cos{ax}$
  
• $(\cos{x})^{(49)}=-\sin{x}\implies (\cos(ax)^{(49)}=-a^{49}\sin{ax}$
• $(\cos{x})^{(48)}=\cos{x}\implies (\cos(ax)^{(48)}=a^{48}\cos{ax}$

Therefore, the final answer is

$$\begin{align*}&x^2(-a^{50}\cos(ax))+ \binom{50}{1}(2x)(-a^{49}\sin(ax))+\binom{50}{2}(2)(a^{48}\cos(ax))\\&=-a^{50}x^2\cos(ax) -100 a^{49}x\sin(a) + 2450a^{48}\cos(ax)\\&=\boxed{a^{48}\left((2450-a^2x^2)\cos(ax)-100ax\sin(ax)\right)}\end{align*}$$