Partial derivatives - verify solution?

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SUMMARY

The discussion centers on the verification of solutions for partial derivatives involving functions f, g, and F defined as F(x,y)=f(x,y,g(x,y)). The solution provided indicates that the total derivative DF can be expressed as DF=D_1F+D_2F=(f_1+f_3g_1)+(f_2+f_3g_2). Furthermore, it concludes that if f_3 is non-zero, the partial derivatives of g can be derived as g_1=-f_1/f_3 and g_2=-f_2/f_3. The participant raises a valid concern regarding the case when f_3=0, noting that no conclusions about the partial derivatives of g can be drawn in that scenario.

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[SOLVED] partial derivatives - verify solution?

Let [tex]f:\mathbb{R}^3\rightarrow\mathbb{R}[/tex], [tex]g:\mathbb{R}^2\rightarrow\mathbb{R}[/tex], and [tex]F:\mathbb{R}^2\rightarrow\mathbb{R}[/tex] be given by
[tex]F(x,y)=f(x,y,g(x,y))[/tex].
1. Find DF in terms of the partial derivatives of f and g.
2. If F(x,y)=0 for all (x,y), find [tex]D_1g[/tex] and [tex]D_2g[/tex] in terms of the partial derivatives of f.

My solution:
1. [tex]DF=D_1F+D_2F=(f_1+f_3g_1)+(f_2+f_3g_2)[/tex]
2. If [tex]f_3\neq0[/tex], then we have the partials of F being zero, so:
[tex]g_1=-f_1/f_3[/tex] and [tex]g_2=-f_2/f_3[/tex]. However, if [tex]f_3=0[/tex] then we have [tex]f_1=f_2=0[/tex].

My concern is with the last part of 2. If [tex]f_3=0[/tex], then I cannot make any statement about the partials of g. Am I doing something wrong?


NOTE: [tex]f_1[/tex] refers to differentiation of f by the first variable.
 
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No, that's completely true. If F is, in fact, NOT a function of g, then no information about F can tell you anything about g!
 
Thanks! :)
 

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