This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Mathematics 105, Spring 2004 — Midterm Exam #1 Solutions (1b) Give an example of a function f : R 3 → R 1 such that f is not differentiable at a = (0 , , 0), but all partial derivatives D i f ( a ) do exist. Solution : Define f (0 , , 0) = 0, and for all x = ( x 1 , x 2 , x 3 ) 6 = (0 , , 0) define f ( x ) = x 1 x 2 x 3 /  x  2 . (Justification was not required, but here it is: All three partial derivatives D i f (0 , , 0) vanish, yet other directional derivatives at 0 = (0 , , 0) are not zero. On the other hand, according to one of our theorems, if f were differentiable at 0 then the Jacobian matrix f (0) would have to equal zero since its entries are the partial derivatives. Now according to a homework problem, all directional derivatives can be calculated from the derivative by D v f (0) = Df (0)( v ). Thus all directional derivatives must also equal 0. But it’s easy to see from the formula that for v = (1 , 1 , 1), say, D v f (0) is 6 = 0.) (2b) Let V, W be two open subsets of R n , and suppose that f : V → W is an invertible function with inverse g . Suppose that f is differentiable at a ∈ V , and g is differentiable at b = f ( a ). State and prove a formula relating f ( a ) to g ( b ). Solution : Let h ( x ) = x for all x ∈ R n . Then g ◦ f ( x ) = h ( x ) for all x ∈ V . Since f, g are differentiable at a and at b = f ( a ), respectively, the chain rule says that h ( a ) = g ( b ) · f ( a ). Now since h is a linear transformation, Dh...
View
Full
Document
This note was uploaded on 05/07/2008 for the course MATH 105 taught by Professor Michaelchrist during the Spring '04 term at University of California, Berkeley.
 Spring '04
 MichaelChrist
 Derivative

Click to edit the document details