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issisoccer10
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[SOLVED] Partial differential with respect to y
Given the equation f(x,y) = (x[tex]^{3}[/tex] + y[tex]^{3}[/tex])[tex]^{1/3}[/tex]
Show that [tex]f_{y}[/tex](0,0) = 1
Basic chain rule..
Based on the chain rule...I believe that
[tex]f_{y}[/tex] = 1/3(x[tex]^{3}[/tex] + y[tex]^{3}[/tex])[tex]^{-2/3}[/tex] 3y[tex]^{2}
or
f_{y}[/tex] = y[tex]^{2}[/tex](x[tex]^{3}[/tex] + y[tex]^{3}[/tex])[tex]^{-2/3}[/tex]
However, plugging in (0,0) does make [tex]f_{y}[/tex] equivalent to 1. Where is my mistake? Any help would be greatly appreciated.. thanks
Homework Statement
Given the equation f(x,y) = (x[tex]^{3}[/tex] + y[tex]^{3}[/tex])[tex]^{1/3}[/tex]
Show that [tex]f_{y}[/tex](0,0) = 1
Homework Equations
Basic chain rule..
The Attempt at a Solution
Based on the chain rule...I believe that
[tex]f_{y}[/tex] = 1/3(x[tex]^{3}[/tex] + y[tex]^{3}[/tex])[tex]^{-2/3}[/tex] 3y[tex]^{2}
or
f_{y}[/tex] = y[tex]^{2}[/tex](x[tex]^{3}[/tex] + y[tex]^{3}[/tex])[tex]^{-2/3}[/tex]
However, plugging in (0,0) does make [tex]f_{y}[/tex] equivalent to 1. Where is my mistake? Any help would be greatly appreciated.. thanks