Question

find the directional derivative of f(x, y) =1+2x√y on point p=(3, 4) in the the direction of u=<4, -3>
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Answer 1

Step-by-step explanation:

f(x,y)=xy∇f(x;y)=⟨fx(x;y),fy(x;y)⟩ fx(x;y)=∂f(x,y)∂x fx(x;y)=∂(xy)∂x fx(x;y)=1y fy(x;y)=∂f(x,y)∂y fy(x;y)=∂(xy)∂y fy(x;y)=−xy2 ⟹∇f(x;y)=⟨1y,−xy2⟩f(x,y)=xy∇f(x;y)=⟨fx(x;y),fy(x;y)⟩ fx(x;y)=∂f(x,y)∂x fx(x;y)=∂(xy)∂x fx(x;y)=1y fy(x;y)=∂f(x,y)∂y fy(x;y)=∂(xy)∂y fy(x;y)=−xy2 ⟹∇f(x;y)=⟨1y,−xy2⟩

 Part (b). Evaluate the gradient at the point P. Part (b). Evaluate the gradient at the point P.

P=(−1,−4)∇f(x;y)=⟨1y,−xy2⟩⟹∇f(x;y)(−1,−4)=⟨−14,116⟩P=(−1,−4)∇f(x;y)=⟨1y,−xy2⟩⟹∇f(x;y)(−1,−4)=⟨−14,116⟩