- #1
wezzo62
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Find ∂fxand ∂fy for: f(x,y)=(3x2+y2+2xy)1/2
i tried using the chain rule and said f(x,y) = u1/2 then
∂fu = 1/2u-1/2, ∂ux = 6x + 2 and ∂uy=2y+2
∂fx = 1/2(3x2 + y2 + 2xy)-1/2(6x + 2)
∂fy = 1/2(3x2 + y2 + 2xy)-1/2(2y + 2)
im not sure if this is correct as i don't know if I am doing the ∂ux and ∂uy part right?
I also have a similar problem with f(x,y)=(x+y)3:
∂fu=3u2, ∂ux and ∂uy=1
in which case both ∂fx and ∂fy=3(x+y)2. . . .
im not sure if ∂ux should be 1+y, ∂uy = x+1 therefore
∂fx=3(x+y)2(1+y) and ∂fy=3(x+y)2(x+1)
any help much appreciated, thanks
i tried using the chain rule and said f(x,y) = u1/2 then
∂fu = 1/2u-1/2, ∂ux = 6x + 2 and ∂uy=2y+2
∂fx = 1/2(3x2 + y2 + 2xy)-1/2(6x + 2)
∂fy = 1/2(3x2 + y2 + 2xy)-1/2(2y + 2)
im not sure if this is correct as i don't know if I am doing the ∂ux and ∂uy part right?
I also have a similar problem with f(x,y)=(x+y)3:
∂fu=3u2, ∂ux and ∂uy=1
in which case both ∂fx and ∂fy=3(x+y)2. . . .
im not sure if ∂ux should be 1+y, ∂uy = x+1 therefore
∂fx=3(x+y)2(1+y) and ∂fy=3(x+y)2(x+1)
any help much appreciated, thanks
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