Question

Find the nth derivative of sin⁴x

Answer 1

Answer:

sin⁴x = ( (1−cos(2x)/2 )² = (1−2cos(2x)+cos²(2x))/4

= ( 1−2cos(2x)+(1+cos(4x))/2 )/4 = ( 3–4cos(2x)+cos(4x) )/8 … (i)

The derivatives of cos(2x) are

n=0 +2⁰cos(2x) = 2⁰cos(2x+0*π/2)

n=1 −2¹sin(2x) = 2¹cos(2x+1*π/2)

n=2 −2²cos(2x) = 2²cos(2x+2*π/2)

n=3 +2³sin(2x) = 2³cos(2x+3*π/2)

n=4 +2⁴cos(2x) = 2⁴cos(2x+4*pi/2)

which generalizes to 2ⁿcos(2x+nπ/2)

Similarly the n’th derivative of cos(4x) is 4ⁿcos(4x+nπ/2)

Applying these to (i) gives for n>0

dⁿ(sin⁴x)/dxⁿ = −½*2ⁿcos(2x+nπ/2) + ⅛*4ⁿcos(4x+nπ/2)