tolong di jawab no 2...please mohon

Kelas 11 Matematika
Bab Turunan

^ = pangkat

2] f(x) = √(2x^5 - 6x + 1)
f(x) = (2x^5 - 6x + 1)^(1/2)
f' (x) = (1/2) . (2x^5 - 6x + 1)^(-1) . (10x^4 - 6)
f' (x) = (5x^4 - 3)/(√(2x^5 - 6x + 1))

f(x) = √(2x⁵ - 6x + 1)

f '(x) = (1/2) . (2x⁵ - 6x + 1)^(-1/2) . (10x⁴ - 6)
f '(x) = (5x⁴ - 3) / √(2x⁵ - 6x + 1)

[tex] f'(x) = \frac{5x^4 -3}{ \sqrt{2x^5 - 6x + 1}} [/tex]