f(x)=(3x^2+4)^5.(2x-1)^4=f¹(x)

Turunan

f(x) = (3x² +4)⁵ (2x -1)⁴

y = uv
y' = u'v + u v'

u = (3x²+4)⁵ --> u' = 5(6x)(3x²+4)⁴ = 30x(3x²+4)⁴
v = (2x-1)⁴ ---> v' = 4(2)(2x-1)³ = 8(2x-1)³

y' = u' v + u v'
y' = 30x(3x²+4)⁴ (2x-1)⁴ + 8(2x-1)³ (3x²+4)⁵

y' = 2(3x²+4)⁴(2x-1)³ [ 15x (2x-1) + 4(3x²+4) ]
y' = 2(3x²+4)⁴(2x-1)³ [ 30x² -15x + 12x² +16 ]
y' = 2(3x²+4)⁴ (2x-1)³ [ 42x² -15x + 16 ]