Question

de as seguintes derivadas ;
1) y=x^3-2x^2+3x
2) y=x+x^3/2-1/3
3) y=(x^2+〖3x)〗^2

Answer 1

Para calcular derivada, vamos usar a regra do tombo ou do expoente, onde:

$f(x)=x^n\rightarrow\ f(x)'=n*x^n^-^1$

Então, temos que:

1)

$y=x^3-2x^2+3x\\\\ \boxed{y'=3x^2-4x+3}$

2)

$y=x+\frac{x^3}{2}-\frac{1}{3}\\\\ y'=1+\frac{3x^2}{2}+0\\\\ \boxed{y'=\frac{3x^2}{2}+1}$

3)

$y=(x^2+(3x)^2\\\\ y'=2x+6x\\\\ \boxed{y'=8x}$