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  • Let $f: \mathbf{R}^{3} \rightarrow \mathbf{R}$ and $g: \mathbf{R}^{2} \rightarrow \mathbf{R}$ be differentiable. Let $F: \mathbf{R}^{2} \rightarrow \mathbf{R}$ be defined by the equation $$ F(x, y)=f(x, y, g(x, y)) $$ Find $D F$ in terms of the partials of $f$ and $g .$

My Attempt. Consider $x=p$, $y=k$ and $r=g(x,y)$. So

$D F=\left[\begin{array}{lll}{\dfrac{\partial p} {\partial x}} \\ {\dfrac{\partial k} {\partial y}} \\ {\dfrac {\partial r}{\partial g}}\end{array}\right]=\left[\begin{array}{lll}{1} \\ {1} \\ {\dfrac {\partial r}{\partial g}}\end{array}\right]$

Please, may you check my attempt and how can I find $\dfrac{\partial r} {\partial g}$, may you help?

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  • $\begingroup$ I added an answer below, may you check? $\endgroup$
    – user295645
    Nov 8, 2019 at 12:46
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    $\begingroup$ checked . . . . ! $\endgroup$
    – janmarqz
    Nov 8, 2019 at 18:19

2 Answers 2

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Make $\phi(x,y)= \left(\begin{array}{c} x\\ y\\ g(x,y) \end{array}\right)$ and consider $F(x,y)=f\circ\phi(x,y)$.

Then by the chain's rule you have $F'=f'\cdot\phi'$ that is \begin{eqnarray*} {\rm grad}F&=&\left[\frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial z}\right]\left(\begin{array}{cc} 1&0\\ 0&1\\ \frac{\partial g}{\partial x}&\frac{\partial g}{\partial y} \end{array}\right),\\ &=&\left[\frac{\partial f}{\partial x}+\frac{\partial f}{\partial z}\frac{\partial g}{\partial x}\ ,\ \frac{\partial f}{\partial y}+\frac{\partial f}{\partial z}\frac{\partial g}{\partial y} \right] \end{eqnarray*} then you will be through.

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  • $\begingroup$ Thanks but I couldn't understand your answer. How did you make $\phi(x,y)$ and why $F(x,y)$ equal to $f\circ \phi (x,y)$? May you explain? $\endgroup$
    – user295645
    Nov 8, 2019 at 12:33
  • $\begingroup$ the choice for $\phi$ is natural because $f\circ\phi(x,y,g)=f(x,y,g)$ $\endgroup$
    – janmarqz
    Nov 8, 2019 at 13:20
  • $\begingroup$ Why did you write comma in your last equality? $DF$ should be$1\times 2$ matrix. $\endgroup$
    – user295645
    Nov 8, 2019 at 14:21
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    $\begingroup$ and for a row matrix any one would use the format [a, b] necessarily XD $\endgroup$
    – janmarqz
    Nov 8, 2019 at 21:01
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    $\begingroup$ hah, okey thanks :) $\endgroup$
    – user295645
    Nov 8, 2019 at 21:58
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By the chain rule

$$F_x=[f(x,y,g(x,y)]_x =f_x(x,y,g(x,y))+f_z(x,y,g(x,y)) g_x(x,y),$$

$$F_y=[f(x,y,g(x,y)]_y =f_x(x,y,g(x,y))+f_z(x,y,g(x,y)) g_y(x,y).$$

Hence,

$$DF= \left[\frac{\partial f}{\partial x}+\frac{\partial f}{\partial z} \frac{\partial g}{\partial x} , \frac{\partial f}{\partial y}+\frac{\partial f}{\partial z} \frac{\partial g}{\partial y}\right]_{1\times 2} $$

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  • $\begingroup$ if you want your try could be a good answer, you ought to explain where those formulas, $F_x$ and $F_y$, come $\endgroup$
    – janmarqz
    Nov 8, 2019 at 18:15

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