How do you find the nth derivative of #f(x) = sin^4(x) + sin^2(x)cos^2(x)#?

1 Answer
Aug 14, 2015

Answer:

# f^n(x) = 2^(n-1)(delta_0 + cos (2x + ((n-2)pi)/2)) " " n in ZZ_0^+ #

where # delta_0 = {""_(0, " otherwise")^( 1, " " n = 0)#

Explanation:

# sin^2 A + cos^2 A = 1 #
# sin A = cos (A-pi/2) #
# cos 2A = 1 - 2sin^2 A #

# f(x) = sin^4 x + sin^2 x cos^2 x = sin^2 x(sin^2 x + cos^2 x) = 1/2(1-cos 2x) = 1/2(1+cos(2x-pi)) #

# f'(x) = sin 2x = cos (2x-pi/2) #
# f''(x) = 2cos 2x = 2cos (2x+0pi) #
# f'''(x) = -4sin 2x = -4 cos (2x-pi/2) = 4 cos (2x+(pi)/2) #
# f''''(x) = -8cos 2x = 8 cos (2x+pi) #

# f^n(x) = 2^(n-1)(delta_0 + cos (2x + ((n-2)pi)/2)) " " n in ZZ_0^+ #

where # delta_0 = {""_(0, " otherwise")^( 1, " " n = 0)#