# Exercises Chapter 2, Section 1

1. Prove that if is differentiable at , then it is continuous at .

If is differentiable at , then . So, we need only show that , but this follows immediately from Problem 1-10.

2. A function is said to be independent of the second variable if for each we have for all . Show that is independent of the second variable if and only if there is a function such that . What is in terms of ?

The first assertion is trivial: If is independent of the second variable, you can let be defined by . Conversely, if , then .

If is independent of the second variable, then because:

Note: Actually, is the Jacobian, i.e. a 1 x 2 matrix. So, it would be more proper to say that , but I will often confound with , even though one is a linear transformation and the other is a matrix.

3. Define when a function is independent of the first variable and find for such . Which functions are independent of the first variable and also of the second variable?

The function is independent of the first variable if and only if for all . Just as before, this is equivalent to their being a function such that for all . An argument similar to that of the previous problem shows that .

4. Let be a continuous real-valued function on the unit circle such that and . define by

1. If and is defined by , show that is differentiable.

One has when and otherwise. In both cases, is linear and hence differentiable.

2. Show that is not differentiable at (0, 0) unless .

Suppose is differentiable at with, say, . Then one must have: . But and so . Similarly, one gets . More generally, using the definition of derivative, we get for fixed : . But , and so we see that this just says that for all . Thus is identically zero.

5. Let be defined by

Show that is a function of the kind considered in Problem 2-4, so that is not differentiable at .

Define by for all . Then it is trivial to show that satisfies all the properties of Problem 2-4 and that the function obtained from this is as in the statement of this problem.

6. Let be defined by . Show that is not differentiable at 0.

Just as in the proof of Problem 2-4, one can show that, if were differentiable at 0, then would be the zero map. On the other hand, by approaching zero along the 45 degree line in the first quadrant, one would then have: in spite of the fact that the limit is clearly 1.

7. Let be a function such that . Show that is differentiable at 0.

In fact, by the squeeze principle using .

8. Let . Prove that is differentiable at if and only if and are, and in this case

Suppose that . Then one has the inequality: . So, by the squeeze principle, must be differentiable at with .

On the other hand, if the are differentiable at , then use the inequality derived from Problem 1-1: and the squeeze principle to conclude that is differentiable at with the desired derivative.

9. Two functions are equal up to order at if

1. Show that is differentiable at if and only if there is a function of the form such that and are equal up to first order at .

If is differentiable at , then the function works by the definition of derivative.

The converse is not true. Indeed, you can change the value of at without changing whether or not and are equal up to first order. But clearly changing the value of at changes whether or not is differentiable at .

To make the converse true, add the assumption that be continuous at : If there is a of the specified form with and equal up to first order, then . Multiplying this by , we see that . Since is continuous, this means that . But then the condition is equivalent to the assertion that is differentiable at with .

2. If exist, show that and the function defined by

are equal up to order at a.

Apply L'Hôpital's Rule n - 1 times to the limit

to see that the value of the limit is . On the other hand, one has:

Subtracting these two results gives shows that and are equal up to order at .