Answer to Question #141182 in Calculus for Faysal

Question

If f(x,y,z)=(x2y2−2y3z2−4x3z2+xyz) , the gradient of f at (1,−2,−1) is

Answer 1

$f(x,y,z) = x^2y^2 - 2y^3z^2 - 4 x^3z^2 + xyz$

$grad f = (\frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial z})$

$\frac{\partial f}{\partial x}= 2xy^2 - 12 x^2 z^2 + yz\\ \frac{\partial f}{\partial y} = 2x^2y - 6y^2z^2 + xz\\ \frac{\partial f}{\partial z} = -4y^3z-8x^3z + xy$

$gradf = (2xy^2 - 12 x^2 z^2 + yz ,\\2x^2y - 6y^2z^2 + xz,\\ -4y^3z-8x^3z + xy )$

And then put $(1,-2,-1)$ into $gradf$

$grad f (1,-2,-1) = (-2,-29,-25)$