Question

alguem sabe resolver derivadas?

Answer 1

Oi :)
Segue a resposta

a)
f(x)= (2x²-1)³
f'(x)=3(2x²-1)².(2x²-1)'
f'(x)=3(2x²-1)².(4x)
f'(x)=12x(2x²-1)²

b)
f(x)= (2x-1)⁴
f'(x)=4(2x-1)³(2x-1)'
f'(x)=4(2x-1)³(2)
f'(x)= 8(2x-1)³

c)
f(x)=(5x²-3x+5)⁶
f'(x)=6(5x²-3x+5)⁵.(5x²-3x+5)'
f'(x)=6(5x²-3x+5)⁵.(10x-3)

h)
f(x)=sen(x²-3x)
f'(x)=cos(x²-3x).(x²-3x)'
f'(x)=cos(x²-3x).(2x-3)
f'(x)=(2x-3)cos(x²-3x)

e assim sucessivamente......