Question 984373
Hi dear,
To differentiate this type of fuction,use product rule
which states that
<prev>
{{{dy/dx=u(dv/dx)+v(du/dx)}}}
let u=x^2--------->du/dx=2x
    v=3x+1-------->dv/dx=3
put the above into the formula
{{{dy/dx=(x^(2)*3)+(3x+1)*2x}}}
dy/dx=3x^2+6x^2+2x
dy/dx=9x^2+2x
HOPE THIS HELPS
Best regards,
Timnewman::