Answer:
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Explanation:
We need to find the directional derivative of the function at the given point in the direction of the vector v.
,point (0, 0, 0) and
By Theorem: If f is a differentiable function of x , y and z , then f has a directional derivative for any unit vector
and
where
since,
then
The partial derivatives are
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
Then the directional derivative is

so, directional derivative at point (0,0,0)


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