Question

me ajudemmmm por favooor ​

Answer 1

Resposta:

1)

Se for  y =x²/[2x√x]=x/[2√x]=√x/2

y'=(1/2)* (1/2) * x^(1/2-1)

y'= 1/(4√x)

2)

y=x^(1/4 +1/3)

y=x^(7/12)

y'=(7/12)* x^(7/12-1)

y'=(7/12) * x^(-5/12)

y'=7/[12*x^(5/12)]

3)

y=x^(1/7) * x^(1/4) * x¹

y=x^(1/7+1/4+1)

y=x^(4/28+7/28+28/28)

y=x^(39/28)

y'=(39/28) * x^(39/28-1)

y'=(39/28)*x^(11/28)