f(x) = ³√(2x² + 6x + 1)²
= (2x² + 6x + 1)^⅔
f'(x) = ⅔ (2x² + 6x + 1)^(⅔ - 1) (4x + 6)
= ⅔(4x + 3) (2x² + 6x + 1)^-⅓
= 2(4x + 3) / 3(2x² + 6x + 1)^⅓
= (8x + 6) / 3 ³√(2x² + 6x + 1)
Mapel : Matematika
Bab : Turunan fungsi aljabar
Kode : 11.2.9
" Life is not a problem to be solved but a reality to be experienced! "
© Copyright 2013 - 2025 KUDO.TIPS - All rights reserved.
f(x) = ³√(2x² + 6x + 1)²
= (2x² + 6x + 1)^⅔
f'(x) = ⅔ (2x² + 6x + 1)^(⅔ - 1) (4x + 6)
= ⅔(4x + 3) (2x² + 6x + 1)^-⅓
= 2(4x + 3) / 3(2x² + 6x + 1)^⅓
= (8x + 6) / 3 ³√(2x² + 6x + 1)
Mapel : Matematika
Bab : Turunan fungsi aljabar
Kode : 11.2.9