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Question Number 192398 by Mastermind last updated on 16/May/23

$$\mathrm{Let}\mathrm{f}:{\mathbb{R}}^3\to \mathbb{R}\mathrm{be}\mathrm{define}\mathrm{by}$$$$\mathrm{f}(\mathrm{x},\mathrm{y},\mathrm{z})={2\mathrm{x}}^2-\mathrm{y}+6\mathrm{xy}-{\mathrm{z}}^3+3\mathrm{z}.$$$$\mathrm{calculate}\mathrm{the}\mathrm{directional}\mathrm{deriva}-$$$$\mathrm{tive}\mathrm{of}\mathrm{the}\mathrm{vector}\mathrm{u}=(2,1,-3)$$$$$$$$\mathrm{help}!$$

Answered by aleks041103 last updated on 21/May/23

$$(u\cdot ▽)f=2{\partial }_{x}f+{\partial }_{y}f-3{\partial }_{z}f=$$$$=2(4x+6y)+(-1+6x)-3(-3z^2+3)=$$$$=8x+12y-1+6x+9z^2-9=$$$$=14x+12y+9z^2-10$$

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