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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...
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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