Put them is ascending order.

Count them = n.

If n is odd, calculate (n+1)/2

the median is the value of the [(n+1)/2]th number in the ordered list.

If n is even, the median is the average of the [n/2]th and [n/2 + 1]th numbers.

Unfortunately, the browser used for posting questions is hopelessly inadequate for mathematics: it strips away most symbols. All that we can see is "...find Th and Th derivative of y - five cos (twox) ...". From that it is not at all clear what the missing symbols (operators) might be. I am guessing that you want the nth and (n+1)th derivatives of something like y = -5*cos(2x). I am also assuming that you are familiar with the chain rule for finding derivatives. Finally, this browser is also rubbish and I cannot show superscripts so I will use the ^ symbol to represent powers..
y = -5*cos(2x)dy/dx = -5*[-sin(2x)]*2 = 2*5*sin(2x) d^2y/dx^2 = 2*5*cos(2x)*2 = 2^2*5*cos(2x)d^3y/dx^3 = 2^2*5*[-sin(2x)]*2 = -2^3*5*sin(2x) d^4y/dx^4 = -2^3*5*cos(2x)*2 = -2^4*5*cos(2x) = 2^4*y

So,for n = 1, 5, 9 ... [n = 1 mod(4)]: d^ny/dx^n = 2^n*5*sin(2x) = 2^n*5*cos(pi/2 - 2x)for n = 2, 6, 10 ... [n = 2 mod(4)]: d^ny/dx^n = 2^n*5*cos(2x) = -2^n*y
for n = 3, 7, 11 ... [n = 3 mod(4)]: d^ny/dx^n = -2^n*5*sin(2x) = -2^n*5*cos(pi/2 - 2x)
for n = 4, 8, 12 ... [n = 0 mod(4)]: d^ny/dx^n = -2^n*5*cos(2x) = 2^n*y

2 n in a d is 2 = nickels in a dime.

'Firme(n)/firma(d) este papel'. (formal/informal(plural))

(FEARmay(n)/FEARmah(th) ESStay pahPELL) ('th' as in 'they')

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