# Derivative of vector-valued function.

1. Jul 7, 2009

### minderbinder

1. The problem statement, all variables and given/known data

Consider the mapping $$f: R^3 \rightarrow R^2$$ defined by

$$f (x_1, x_2, x_3) = (x_1 e^{x_2}, x_2 e^{x_3})$$.

i) Show that the mapping $$f(x)$$ is differentiable and find its derivative at the origin.
ii) Show that the equation $$f(x) = a$$ for any point $$a \in R^2$$ determines a one-dimensional submanifold $$C \subset R^2$$, a smooth curve C in the three-dimensional space that can be described locally by giving two of the three variables $$x_1, x_2, x_3$$ as functions of the other variable.
iii) Find a tangent vector to the curve $$f(x) = (e, e)$$ at the point (1, 1, 1).

2. Relevant equations

I'm studying on my own, and unfortunately no solution is provided for this problem. I've got everything pretty much figured out, and only have some small questions. Can someone please look over my work & let me know anything I'm doing wrong? Thanks.

3. The attempt at a solution

i)

The derivative matrix is:
[$$e^{x_2} x_1 e^{x_2} 0$$]
[$$0 e^{x_3} x_2 e^{x_3}$$]

The existance of this matrix shows that $$f(x)$$ is differentiable. (Is my reasoning here valid? Or would I need to do more to show that the function is valid?)

The derivative matrix at the origin is:
[1 0 0]
[0 1 0]

ii)
We have:
$$x_1 e^{x_2} = a_1$$ and $$x_2 e^{x_3} = a_2$$. Adding them gives $$x_1 e^{x_2} + x_2 e^{x_3} - (a_1 + a_2) = 0$$, and we let this equation be called $$g(x)$$. $$\nabla g(x) = (e^{x_2}, x_1 e^{x_2} + e^{x_3}, x_2 e^{x_3})$$. By the implicit function theorem, since $$\nabla g(x)$$ is never zero (because $$e^{x_2}$$ is never zero), this curve can be solved locally anywhere. I'm unsure about how to show that this is a curve instead of a surface...

iii) Plug in (1, 1, 1) into $$\nabla g(x)$$, we have (e, 2e, e).