How to find the 48th derivative of the function cos2x?NB: Is there any other possible way without having to do each derivative from 1-48?

Let us find the series of differentiating
cos2x.

==> Let y=
cos2x

=> y' = -2sin2x

==>
y'' = -2*2cos2x = -4cos2x

==> y''' = -4*-2*sin2x =
8sin2x

==> y'''' = 8*2*cos2x =
16cos2x

Now we will find the relation of the
sequence.

==> -2sin2x, -4cos2x, 8sin2x, 16cosx , -32sin2x,
-64cos2x,....

an : -2sin2x, -32sin2x, ..... n= 1, 5, 9,
...

an: -4cos2x , -64cos2x, ... n= 2, 6, 10,
...

an: 8sin2x , 128sin2x, .... n= 3, 7, 11,
...

an : 16cos2x, 256cos2x ,.. n= 4, 8, 12,
...

Now we notice that the 4th derivative belongs to the 4th series
which is an: 16cos2x , 256co2x n= 4, 8, 12,16, 20, ....

==>
a48 = (2)^48 *cos2x

==>Then the 48th
derivative is : 2^48 * cos2x