Finding Gradient Vector of f(x,y,z) = 2*sqrt(xyz) at Point (3,-4,-3)

[kasse]

I want to find the gradient vector of f(x,y,z)=2*sqrt(xyz) at the point ((3,-4,-3).

I find the partials and set in for the x-, y-, and z-values, and find the grad. vector (2, (1,5), 2). The right solution is (2, (-1,5), -2), so I have obviously made a mistake with the sqrt. How do I know whether it's + or -?

And another question: Have I used the chain rule correctly when I have calculated the gradient of f(x,y,z) at (-5,1,3) to be (160,-240,400)?

 

[stunner5000pt]

what did you get for your gradient??

 

[arildno]

Call your variables [itex]x_{i}, i=1,2,3[/tex]<br /> Then, $f(x_{1},x_{2},x_{3})=2\sqrt{x_{1}x_{2}x_{3}}$<br /> And the partials are:<br /> $$\frac{\partial{f}}{\partial{x}_{i}}=\frac{2x_{j}x_{k}}{f}, i,j,k=1,2,3,j\neq{i}\neq{k},j\neq{k}$$<br /> Mind your signs..[/itex]

 

[HallsofIvy]

I want to find the gradient vector of f(x,y,z)=2*sqrt(xyz) at the point ((3,-4,-3).

I find the partials and set in for the x-, y-, and z-values, and find the grad. vector (2, (1,5), 2). The right solution is (2, (-1,5), -2), so I have obviously made a mistake with the sqrt. How do I know whether it's + or -?

And another question: Have I used the chain rule correctly when I have calculated the gradient of f(x,y,z) at (-5,1,3) to be (160,-240,400)?

?? You know whether it is + or - (- is correct here) because that's what you get when you do the arithmetic!

What did you get for the gradient vector of $f(x,y,z)= 2\sqrt{xyz}$?