nth derivative

applying Leibniz Rule:
y^(n)
= x²(cos x)^(n) + nC1 (2x)(cos x)^(n-1) + nC2 (2)(cos x)^(n-2)
= x²(cos x)^(n) + 2nx(cos x)^(n-1) + n(n-1)(cos x)^(n-2)
where ^(n) denotes the nth derivative, ^(n-1) denotes the (n-1)st derivative etc

for n= 4k,
y^(n) = x²(cos x) + 2nx(sin x) - n(n-1)(cos x)

for n=4k+1
y^(n) = -x²(sin x) + 2nx(cos x) + n(n-1)(sin x)

for n = 4k+2
y^(n) = -x²(cos x) - 2nx(sin x) + n(n-1)(cos x)

for n=4k+3
y^(n) = x²(sin x) - 2nx(cos x) - n(n-1)(sin x)

for all non-negative integer k