Question

помощь не помешает 3 и 4​

Answer 1

(f*g)'=f'*g + f*g'

3)

((x+2)*∛x)'=(x+2)'*∛x + (x+2)*(∛x)'=

1*∛x + (x+2)*(1/3)*x^(1/3 - 1)

∛x + 1/3(x+2)*x^(-2/3)=

∛x + (x+2)/(3*∛x²).

4)

((x-1)√x)'=(x-1)'*√x + (x-1)*(√x)'=

1*√x + (x-1)*(1/2)*x^(1/2-1)=

√x + 1/2 * (x-1) * x^(-1/2)=

√x + (x-1)/(2√x).

______________________________

√x=x^(1/2);  ∛x=x^(1/3).

x^(-2/3)=1/x^(2/3)=1/∛x²;  x^(-1/2)=1/x^(1/2)=1/√x.