Past exam of the options I for mathematics and philosophy course of the University of Oxford / 2013 / F / 1 / a / ii

Table of contents
- Solution ii

Solution

The denominator is positive on

B

because

x_{3} > - 1

. The quotient rule gives the gradient

\nabla g (x) = (\frac{1}{x _{3} + 3}, \frac{1}{x _{3} + 3}, - \frac{x _{1} + x _{2}}{( x _{3} + 3 ) ^{2}}) .

In particular

g

is continuously differentiable throughout

B

.

 Ancestors (13)

 View article source

 Discussion (0) New discussion

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version