# Find the maximum value of the directional derivative of f(x, y, z)=x^2+y^2+z^2-3xy+2xz-yz at the point (1, 2, -1).

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You need to find the gradient of the function such that:

`grad f(x,y,z) = <(del f)/(del x),(del f)/(del y),(del f)/(del z)>`

`(del f)/(del x) = 2x + 0 + 0 - 3y + 2z - 0 `

`(del f)/(del x) = 2x - 3y + 2z`

`(del f)/(del y) = 0 + 2y + 0 - 3x + 0 - z`

`(del f)/(del y)= 2y - 3x - z`

`(del f)/(del z) = 0 + 0 + 2z - 0 + 2x - y`

`(del f)/(del z) = 2z + 2x - y`

You need to evaluate directional derivative at `(1,2,-1)`

`(del f)/(del x)|_(1,2,-1) =2 - 3*2 + 2*(-1)`

`(del f)/(del x)|_(1,2,-1) = -6`

`(del f)/(del y)|_(1,2,-1) =2*2 - 3 - (-1) = 2`

`(del f)/(del z)|_(1,2,-1) =2(-1) + 2 - 2 = -2`

You need to evaluate the maximum value of directional derivative at `(1,2,-1)` , such that:

`grad f(1,2,-1) = sqrt((-6)^2 + 2^2 + (-2)^2)`

`grad f(1,2,-1) = sqrt(36 + 4 + 4) = sqrt44 = 2sqrt11`

**Hence, evaluating the maximum value of directional derivative of `f(x,y,z)` , at` (1,2,-1)` , yields `grad f(1,2,-1) = 2sqrt11` .**