Hitunglah nilai f'(phi per 4) dari f(x)=cos x PER sinx+cos x

f(x) = cos x / (sinx + cosx)

misal
u = cos x.............................u' = -sinx
v = sinx + cosx ...................v' = cos x - sinx
v² = sin²x + cos²x + 2 sinx . cosx = 1 + 2 sinx cosx
 
f'(x) = (u' v - u v') / v²
f'(x) = (-sinx . (sinx + cos x) - (cosx (cosx - sinx))/(1 + 2 sinx cosx)
f'(x) = (-sin²x - sinx cosx - cos²x + sinx cosx) / (1 + 2 sinx . cosx)
f'(x) = (-sin²x - cos²x)/(1+ 2 sinx . cosx)
f'(x) = -1(sin² + cos²x)/(1 + 2 sinx cos x)
f'(x) = - 1(1) / 1 + 2 sinx cosx 
f'(x) = -1/ 1 + 2 sin x cos x

π/4 = 180/4 = 45°


f'(π/4) = f(45°)
f'(π/4) = -1/(1 + 2 sin 45 . cos 45)
f'(π/4) = -1/ (1 + 2. ¹/₂√2  ¹/₂√2)
f'(π/4) = -1/ (1 + 1)
f'(π/4) = -1/2

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