$\phi =x^4 + y^4 + z^4$ $\therefore \nabla \phi = 4 x^3 \overrightarrow {i} + 4y^3 \overrightarrow {j} + 4z^3 \overrightarrow {k}$ At point A (1, -2, 1) $\nabla \phi = 4(1)^3 \overrightarrow {i} + 4(-2)^3 \overrightarrow {j} + 4 (1)^3 \overrightarrow {k}$ $= 4\overrightarrow {i}- 32 \overrightarrow {j} + 4\overrightarrow {k}$ Let $\overrightarrow {p} = \overline{AB}$ $\overrightarrow b - \overrightarrow a$ $= (2\overrightarrow {i} + 6 \overrightarrow {j} - 1 \overrightarrow k ) - ( 1\overrightarrow {i} - 2 \overrightarrow j + 1 \overrightarrow {k})$ $= \overrightarrow {i} + 8 \overrightarrow j - 2 \overrightarrow k$ $\displaystyle \therefore |\overrightarrow p| = \sqrt{1^2 + 8^2 + (-2)^2} = \sqrt{69}$ Directional derivative of ϕ in the direction of AB $= \nabla \phi\cdot \dfrac {\overrightarrow p}{|\overrightarrow p|}$ $=4(\overrightarrow i - 32 \overrightarrow j + 4 \overrightarrow k)\cdot \dfrac {(\overrightarrow i + 8 \overrightarrow j - 2 \overrightarrow k)}{\sqrt{69}}$ $= \dfrac {4-256-8}{\sqrt{69}}$ $= \dfrac {-260}{\sqrt{69}} $ Maximum directional derivate of ϕ at (1, -2, 1) $= |\nabla \phi |$ $= \sqrt{4^2 + (-32)^2 + (4)^2}$ $= 4 \sqrt{66}$