- #1

issisoccer10

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**[SOLVED] Partial differential with respect to y**

## 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