Question

Please help with how you calculate directional derivative

In this question, given the function f(x,y,z,w)= [x^2y+zw, y^2z+xw, xyzyw] find the differential df and the directional derivative at [-1,0,1,2]

Answer 1

$f(x,y,z,w) = (x^{2}y + zw,y^{2}z + xw, xyz, yw)$

$df = 2yxdx+2zydy + xydz + ydw$

$\cfrac{\partial f}{\partial l} = \cfrac{\partial f}{\partial x}cos\alpha + \cfrac{\partial f}{\partial y} cos\beta + \cfrac{\partial f}{\partial z}cos\gamma + \cfrac{\partial f}{\partial w} cos\theta$

Find the direction cosines:

$l = (-1;0;1;2)\\ l_0 = \cfrac{l}{|l|} = \cfrac{-i+k+2t}{\sqrt{1+0+1+4}} = -\cfrac{1}{\sqrt{6}}i+\cfrac{1}{\sqrt{6}}k + \cfrac{2}{\sqrt{6}}t$

so we have:

$cos \alpha = -\cfrac{1}{\sqrt{6}},cos\beta = 0, cos\gamma = \cfrac{1}{\sqrt{6}}, cos\theta = \cfrac{2}{\sqrt{6}}$

so directional derivative at $(-1;0;1;2)$ :

$\cfrac{\partial f}{\partial l} = -\cfrac{2yx}{\sqrt{6}} + \cfrac{xy}{\sqrt{6}} + \cfrac{2y}{\sqrt{6}} = -\cfrac{yx+2y}{\sqrt{6}}$