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8. (12 poluia) Find the dimetional derivutim of \( f(x, y, a)=x^{3} a^{3} \) at the point \( (2,1,1) \) in the dinfictos of \( v=i-2 i+k \).

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To find the directional derivative of a three dimensional function f(x, y, a), we first need to compute the gradient of the function, which is a vector that contains partial derivatives of the f with respect to the each variable.

Given f(x, y, a) = x³a³, the partial derivative with respect to x, denoted as df/dx, will be 3x²a³. The partial derivative with respect to y, df/dy, is 0 since there is no y-term in the function. The partial derivative with respect to a, df/da, is 3x³a².

Therefore, the gradient of f, denoted by ∇f, is given by the vector:

∇f = [, , ] = [3x²a³, 0, 3x³a²]

Next, we evaluate the gradient at the point (2, 1, 1). Substituting these values into each component of the gradient vector results in:

∇f(2, 1, 1) = [3*(2²)*(1³), 0, 3*(2³)*(1²)]
= [12, 0, 24]

The final step is to compute the dot product of the gradient at point (2,1,1) with the direction vector v = [1, -2, 1]. The dot product of two vectors is obtained by multiplying their corresponding components and then summing these products:

v . ∇f(2, 1, 1) = [1, -2, 1] . [12, 0, 24]
= 1*12 + -2*0 + 1*24
= 36

Therefore, the directional derivative of f at (2,1,1) in the direction of v is 36.

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