A multivariable chain rule problem

[501622731]

Hello all,

I am stuck on what seems like a rather simple problem:

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

I would greatly appreciate any help.

 

[Tac-Tics]

What work have you done on it so far? Where are you getting stuck?

(Also, if this is homework, we have a separate forum for that).

 

[501622731]

$$\begin{array}{ll} DF(x,y) &= Df(x,y,g(x,y))\\ &= \begin{bmatrix} D_1f(x,y,g(x,y)) & D_2f(x,y,g(x,y)) & D_3f(x,y,g(x,y)) \end{bmatrix} \end{array}$$
I suspect that the chain rule should be used, in some way, to evaluate $$D_3f(x,y,g(x,y))$$.

Also, this isn't homework; I just started reading Analysis on Manifolds by James Munkres a few days ago.

 

[Tac-Tics]

$$\begin{array}{ll} DF(x,y) &= Df(x,y,g(x,y))\\ &= \begin{bmatrix} D_1f(x,y,g(x,y)) & D_2f(x,y,g(x,y)) & D_3f(x,y,g(x,y)) \end{bmatrix} \end{array}$$
I suspect that the chain rule should be used, in some way, to evaluate $$D_3f(x,y,g(x,y))$$.

I think I see where you are getting confused. Hint: There is no such thing as $$D_3F$$ :-o

Since F is a function of $$\mathbb{R}^2$$, $$D_1F = \frac{dF}{dx}$$, and $$D_2F = \frac{dF}{dy}$$. Therefore, the gradient (DF) is going to be $$(D_1F, D_2F)$$.

The notation is misleading. You probably want to write out the definition of the partial derivative, $$D_1 F = \frac{F(x+\epsilon, y) - F(x, y)}{\epsilon}$$, (and similarly for $$D_2 F$$). Only then should you expand F out.