Jawab:

0

Penjelasan dengan langkah-langkah:

Diketahui fungsi f(x):

f(x) = cos² (x + π/4

Ditanya:   f'(π/4)  ?

f(x) = cosⁿ (ax + c)

• Menggunakan rumus turunan

→ f'(x) = n. cosⁿ⁻¹ (ax + c). c . - sin (ax + c)

→ f'(x) = 2 . cos (x + π/4) . 1 . - sin (x + π/4)

→ f'(x) = - 2 . cos (x + π/4) . sin (x + π/4)

Ingat, 2 cos a sin b = sin (a+b) - sin (a-b)

→ f'(x) = sin (x + π/4 + x + π/4) - sin (x + π/4 - x - π/4)

→ f'(x) = sin (2x + π/2)

• Subtitusi x = π/4 ke turunan tadi

→ f'(π/4) = sin (2(π/4) + π/2)

→ f'(π/4) = sin (π/2 + π/2)

→ f'(π/4) = sin π

→ f'(π/4) = sin 180° = 0

Jadi nilai f’(π/4) adalah 0