Below is a problem from Munkre's Analysis on Manifolds book. I'm unsure of how to approach this; it seems to me to apply the defintion of the derivitative, but I cannot seem to get that to work out. Can anyone offer any hints?

If $f \colon \mathbb{R}^3 \to \mathbb{R}$ and $g \colon \mathbb{R}^2 \to \mathbb{R}$ be differentiable functions. Let $F \colon \mathbb{R}^2 \to \mathbb{R}$ be defined by the equation $$ F(x,y) = f(x,y,g(x,y)). $$ (a) Find $DF$ in terms of the partials of $f$ and $g$. [ ] (b) If $F(x,y)=0$ for all $(x,y)$, find $D_1g$ and $D_2g$ in terms of the partials of $f$.


2 Answers 2


$$ F_x = f_x + f_zg_x,\ F_y=f_y+f_zg_y $$

If $F$ is constant from the above we have $$ g_x=-f_x/f_z,\ g_y= -f_y/f_z$$

Addendum - Chain rule : In general we have formula : $F(t)=f(g(t),h(t),i(t))$ Then $$ \frac{d}{dt} F=f_xg'(t) + f_y h'(t) + f_z i'(t) $$

  • $\begingroup$ I honestly don't follow your answer. given the information I have, I see no way to be able to manipulate the defintion in order to get close to any solution at all. $\endgroup$ Mar 20, 2016 at 15:43
  • $\begingroup$ Which point you can not follow ? $\endgroup$
    – HK Lee
    Mar 20, 2016 at 15:46
  • $\begingroup$ Let's say I'm looking for $F_x$. Then I compute the following limit: $$lim_{t\to 0} \frac{f (x+t, y, g (x+t,y))-f (x,y,g (x,y))}{t} $$, which I don't directly see how to do. $\endgroup$ Mar 20, 2016 at 15:49
  • $\begingroup$ Recall that $DF=[F_x\ F_y]$ To find $F_x$, we must differentiate $f(x,y,g(x,y))$ wrt $x$. So by ${\bf chain rule}$ the result is in the above (You're right That is note that the limit is given by the formula chain rule) $\endgroup$
    – HK Lee
    Mar 20, 2016 at 15:50
  • $\begingroup$ I overlooked something so simple. I verified it for myself. Thank you! $\endgroup$ Mar 20, 2016 at 16:06

Let $c(x,y) = (x,y,g(x,y))$. If we wish to get $F_x, F_y$ we use the chain-rule for paths which implies;

$$F_x = \frac{d}{dx} \ f \circ c(x,y) = \langle f_x ,f_y,f_z\rangle|_{c(x,y)} \cdot \langle 1,0,g_x(x,y)\rangle = f_x + f_z \cdot g_x$$

$$F_y = \frac{d}{dy} \ f \circ c(x,y) =\langle f_x ,f_y,f_z\rangle|_{c(x,y)} \cdot \langle 0,1,g_y(x,y)\rangle = f_y + f_z \cdot g_y$$

Therefore the jacobian $DF$ is given by;

$$\begin{bmatrix} f_x+f_z \cdot g_x & \ \ f_y + f_z \cdot g_y \end{bmatrix}$$

If $F(x,y) = 0$ then we have;

$$f_x + f_z \cdot g_x = 0 \Rightarrow g_x = -\frac{f_x}{f_z}$$

$$f_y+f_z \cdot g_y = 0 \Rightarrow g_y = - \frac{f_y}{f_z}$$


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