Question #141182
If f(x,y,z)=(x2y2−2y3z2−4x3z2+xyz) , the gradient of f at (1,−2,−1) is
1
Expert's answer
2020-11-01T14:53:05-0500

f(x,y,z)=x2y22y3z24x3z2+xyzf(x,y,z) = x^2y^2 - 2y^3z^2 - 4 x^3z^2 + xyz

gradf=(fx,fy,fz)grad f = (\frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial z})

fx=2xy212x2z2+yzfy=2x2y6y2z2+xzfz=4y3z8x3z+xy\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=(2xy212x2z2+yz,2x2y6y2z2+xz,4y3z8x3z+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)(1,-2,-1) into gradfgradf


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


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