# A question from Munkres' Analysis on Manifolds (P. 63, E. 3)

• 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?

– user295645
Nov 8, 2019 at 12:46
• checked . . . . ! Nov 8, 2019 at 18:19

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.

• 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?
– user295645
Nov 8, 2019 at 12:33
• the choice for $\phi$ is natural because $f\circ\phi(x,y,g)=f(x,y,g)$ Nov 8, 2019 at 13:20
• Why did you write comma in your last equality? $DF$ should be$1\times 2$ matrix.
– user295645
Nov 8, 2019 at 14:21
• and for a row matrix any one would use the format [a, b] necessarily XD Nov 8, 2019 at 21:01
• hah, okey thanks :)
– user295645
Nov 8, 2019 at 21:58

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,

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

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