Nilai f'( π ) = ….
2
Jawab:
f'(x)= 2(1 + sinx)cosx(1 + cos)⁴ + 4(1 + cosx)(-sinx)(1 + sinx)²
f'(x) = 2 cosx(1 + sinx)(1 +cosx)⁴ – 4 sinx(1 + cosx)(1 + sinx)²
f'(π/2) = 2 cosπ/2(1 + sinπ/2)(1 + cosπ/2)⁴ – 4 sinπ/2(1 + cosπ/2)(1 + sinπ/2)²
f'(π/2) = 2 . 0 – 4(1)(1 + 0)(1 + 1)²
f'(π/2) = -4(2)²
Makara, f'(π/2) = -16