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du = -cos ((π/4) - x) dx
f(x) = u⁴
f'(x) = 4.u³ du
= 4.sin³ ((π/4) - x).(-cos ((π/4) - x))
f"(x) = (4sin³ ((π/4) - x))(-sin ((π/4) - x)) + (12sin² ((π/4) - x))(cos² ((π/4) - x))
= -(4sin⁴ ((π/4) - x)) + 12.(sin² ((π/4) - x)).(cos² ((π/4) - x))
= -(4sin⁴ ((π/4) - x)) + 3.(4.sin² ((π/4) - x).cos² ((π/4) - x)
= -(4sin⁴ ((π/4) - x)) + 3.sin² 2((π/4) - x)
= -(4sin⁴ ((π/4) - x)) + 3.sin² ((π/2) - 2x); sin (90-x) = cos x
= -(4sin⁴ ((π/4) - x)) + 3.cos² 2x.....opsi E