Answer
(a) \[\nabla f(x,y,z)=(2xyz-yz^3)\hat{i}+(x^2z-xz^3)\hat{j}+(x^2y-3xyz^2)\hat{k}\]
(b) \[-3\hat{i}+2\hat{j}+2\hat{k}\]
(c) \[\frac{2}{5}\]
Work Step by Step
It is given that \[f(x,y,z)=x^2yz-xyz^3\]
Differentiate $f$ with respect to $x$ treating $y$ and $z$ constant.
\[f_x(x,y,z)=2xyz-yz^3\]
Differentiate $f$ with respect to $y$ treating $x$ and $z$ constant.
\[f_y(x,y,z)=x^2z-xz^3\]
Differentiate $f$ with respect to $z$ treating $x$ and $y$ constant.
\[f_z(x,y,z)=x^2y-3xyz^2\]
(a) \[\nabla f(x,y,z)=f_x(x,y,z)\hat{i}+f_y(x,y,z)\hat{j}+f_z(x,y,z)\hat{k}\]
\[\nabla f(x,y,z)=(2xyz-yz^3)\hat{i}+(x^2z-xz^3)\hat{j}+(x^2y-3xyz^2)\hat{k}\]
(b) \[\nabla f(2,-1,1)=f_x(2,-1,1)\hat{i}+f_y(2,-1,1)\hat{j}+f_z(2,-1,1)\hat{k}\]
\[\nabla f(2,-1,1)=-3\hat{i}+2\hat{j}+2\hat{k}\]
(c) \[\mathbf{u}=(0)\hat{i}+\frac{4}{5}\hat{j}-\frac{3}{5}\hat{k}\]
Required directional derivative is given by:
\[D_{\mathbf{u}}f(2,-1,1)=\nabla f(2,-1,1)\cdot \mathbf{u}\]
\[D_{\mathbf{u}}f(2,-1,1)=(-3\hat{i}+2\hat{j}+2\hat{k})\cdot \left((0)\hat{i}+\frac{4}{5}\hat{j}-\frac{3}{5}\hat{k}\right)\]
\[D_{\mathbf{u}}f(2,-1,1)=\frac{2}{5}\]
