f(x)=1/√x+1 berapa f'(x)

jika pnyebutnya smua dlm akar pngkat dua dan lambangnya sama f(bukan F)
f(x) = 1/√(x + 1) = (x + 1)^-(1/2)
f'(x) = -(1/2)(x + 1)^-(3/2)
= -1 / [2√(x + 1)³], atau
= -1 / [2(x + 1)√(x + 1)]

f(x) = 1/√x+1
f(x) = (x+1)^-(1/2)

f'(x) = -½ (x+1)^-(3/2)
f'(x) = -½ × 1 / √(x+1)³
f'(x) = -1 / (2 √(x+1)³)
f'(x) = - 1 / (2(x+1)√(x+1))