Can anyone please help with the following equation? (file attached).
The 3xy2 is confusing me.
Thanks
The equation is
$\displaystyle x^3 + 3xy^2 + 4y = 40$
After differentiation you get
$\displaystyle 3x^2 + 3x(2y)(dy/dx) + 3y^2 + 4dy/dx = 0$
Now simplify to find dy/dx
The equation is
$\displaystyle x^3 + 3xy^2 + 4y = 40$
After differentiation you get
$\displaystyle 3x^2 + 3x(2y)(dy/dx) + 3y^2 + 4dy/dx = 0$
Now simplify to find dy/dx