I couldn't find the solution of this question, can you help me, please?
Let the function f(x,y,z) be differentiable at a point P. The maximum value of the directional derivative of f at P is 6 and the direction of v=(1,1,-1). Find the the gradient vector of f at P.
The gradient is the vector pointed in the direction of maximum ascent, and its size is the rate at which the function is ascending. With that definition, you can see that we have all the information we need.
We have the direction of maximum ascent: $$(1,1,-1).$$ A unit vector in that direction is $$(1,1,-1)/\sqrt{3}.$$ And we know the rate of ascent is $6$; therefore, the gradient is $$\left(6/\sqrt{3}\right)(1,1,-1).$$