# Find the directional derivative of F(x,y,z)=x(y^2)-4(x^2)y+(z^2) at point P(1, -1, 2) in the direction of 6i+2j+3k.

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### 1 Answer

You need to verify if the vector `bar u = 6bar i+2bar j+3 bar k` is the unit vector, hence you need to evaluate what its magnitude is such that:

`|bar u| = sqrt(6^2 + 2^2 + 3^2) =gt |bar u| = sqrt 49`

`|bar u| = 7 != 1 =gt` vector u is not the unit vector, hence you need to converting u to unit vector dividing the vector by its magnitude `|bar u|` such that:

`bar v = (6/7)bar i + (2/7)bar j + (3/7)bar k`

You need to remember the formula of directional derivative such that:

`D_(bar u) f(x,y,z) = (6/7)*f_x(x,y,z) + (2/7)*f_y(x,y,z) + (3/7)*f_z(x,y,z)`

You need to evaluate `f_x(x,y,z), ` hence you need to differentiate the function f(x,y,z) with respect to x, keeping y and z as constants such that:

`f_x(x,y,z)= y^2 - 8xy`

You need to evaluate `f_y(x,y,z),` hence you need to differentiate the function f(x,y,z) with respect to y, keeping x and z as constants such that:

`f_y(x,y,z) = 2yx - 4x^2`

You need to evaluate `f_z(x,y,z),` hence you need to differentiate the function `f(x,y,z)` with respect to z, keeping x and y as constants such that:

`f_z(x,y,z) = 2z`

`D_(bar u) f(x,y,z) = (6/7)*(y^2 - 8xy) + (2/7)*(2yx - 4x^2) + (3/7)*(2z)`

You need to evaluate directional derivative at the point `P(1,-1,2), ` hence you need to substitute 1,-1,2 for x,y,z, in equation `D_(bar u) f(x,y,z) = (6/7)*(y^2 - 8xy) + (2/7)*(2yx - 4x^2) + (3/7)*(2z) ` such that:

`D_(bar u) f(1,-1,2) = (6/7)*(1+ 8) + (2/7)*(-2 - 4) + (3/7)*(4)`

`D_(bar u) f(1,-1,2) = 54/7 - 12/7 + 12/7`

`D_(bar u) f(1,-1,2) = 54/7`

**Hence, evaluating directional derivative of `f(x,y,z)=x*y^2-4x^2*y+z^2` at the point `P(1,-1,2)` in the direction `bar u = 6bar i+2bar j+3bar k,` yields `D_(bar u) f(1,-1,2) = 54/7.` **