Answer
$y^{(50)}=-2^{50}cos(2x)$
Work Step by Step
With trig functions, we can find a pattern with derivatives. Let's start with finding the first five derivatives.
$y=cos(2x)$
$y'=-2sin(2x)$
$y''=-4cos(2x)$
$y'''=8sin(2x)$
$y^{(4)}=16cos(2x)$
$y^{(5)}=-32sin(2x)$
The first thing to note is that for every even numbered derivative, we have a $cos(2x)$ and for every odd numbered derivative, we have a $sin(2x)$.
The second thing to note is the constant in front of the trig function. The pattern is that for the $n^{th}$ derivative, the constant is $2^{n}$.
The last thing to note is the sign. This is a little trickier because it changes every two derivatives. We are looking at the $50^{th}$ derivative, and we know that $\frac{50}{4}=12$ with a remainder of $2$. If we look at the second derivative we found earlier ($y''$), we see that the sign is negative ($-$).
If we put all three notes together, we see that our derivative is negative ($-$), has a constant of $2^{50}$, and is multiplied by $cos(2x)$.
As a result, $y^{50}=-2^{50}cos(2x)$.
