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}<br /> DF(x,y) &amp;= Df(x,y,g(x,y))\\<br /> &amp;= \begin{bmatrix} D_1f(x,y,g(x,y)) &amp; D_2f(x,y,g(x,y)) &amp; D_3f(x,y,g(x,y)) \end{bmatrix}<br /> \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}<br /> DF(x,y) &amp;= Df(x,y,g(x,y))\\<br /> &amp;= \begin{bmatrix} D_1f(x,y,g(x,y)) &amp; D_2f(x,y,g(x,y)) &amp; D_3f(x,y,g(x,y)) \end{bmatrix}<br /> \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.