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So in order to understand the general solution here, let's start with some simple cases, some simple cases. So if N is equal to one, then there are only two possibilities here. We either have, well, the partial derivative with respect to X. So if some X or the powerful derivative with respect to Y f sub y. Um, if n equals two, well, then there are four possibilities.

(b) If these partial derivatives are all continuous, how many of them can be distinct? (c) Answer the question in part (a) for a function of three ...

(b) If these partial derivatives are all continuous, how many of them can be distinct? (c) Answer the question in part (a) for a function of five variables.

5.Assume f and all of its rst, second and third partial derivatives are continuous. How many possible third partials are there? How many can be distinct? 6. Optional: How many nth-order partial derivatives does a function of two variables have? If they’re all continuous, how many of them can be distinct?

If these partial derivatives are all continuous, how many of them can be distinct? Solution. Counting mixed partials as different, we get 2 kor n for 2 or n variables, respectively. If they are all continuous, then for the case of 2 variables, we only care how many times we take derivative with respect to x, which could be any integer

(b) If these partial derivatives are all continuous, how many of them can be distinct? (c) Answer the question in part (a) for a function of three ...

(b) If these partial derivatives are all continuous how many of them can be distinct? 2n Answer the question in part (2) for function of three variables.

Well, this is clearly to to the end. So we have to to the end party here is just to to the end, um, part B, it says, Um, well, if these parts derivatives are all continuous, how many of them can be distinct or that would be then, um, that would be m plus one. That would be m plus one part B. But then, um, party, right? If we have three, very ...

This is the problem of distributing n balls over k bins, which can be solved using the stars and bars approach; the result is.

This problem has been solved! (a) How many n th-order partial derivatives does a function of two variables have? (b) If these partial derivatives are all continuous, how many of them can be distinct? (c) Answer the question in part (a) for a function of five variables. This is the best answer based on feedback and ratings.

Each of these partial derivatives is a function of two variables, so we can calculate partial derivatives of these functions. Just as with derivatives of single-variable functions, we can call these second-order derivatives, third-order derivatives, and so on. In general, they are referred to as higher-order partial derivatives. There are four ...

78E. 79E. (a) How many n th-order partial derivatives does a function of two variables have? (b) If these partial derivatives are all continuous, how many of them can be distinct? Step-by-step solution. Step 1 of 3. (A) For a function f of two variables the …

Problem 98E. (a) How many n th-order partial derivatives does a function of two variables have? (b) If these partial derivatives are all continuous, how many of them can be distinct? (c) Answer the question in part (a) for a function of three variables. Chapter 14.3, Problem 98E is solved.

Partial derivatives and continuity. Theorem If the function f : R → R is differentiable, then f is continuous. Remark: I This Theorem is not true for the partial derivatives of a function f : R2 → R. I There exist functions f : R2 → R such that f x(x 0,y 0) and f y (x 0,y 0) exist but f is not continuous at (x 0,y 0). 1 f(x,y) C C 1 2 x ...

If these partial derivatives are all continuous, how many of them can be distinct? Solution. Counting mixed partials as different, we get 2 kor n for 2 or n variables, respectively. If they are all continuous, then for the case of 2 variables, we only care how many times we take derivative with respect to x, which could be any integer

12.01.2021 · So we have to to the end party here is just to to the end, um, part B, it says, Um, well, if these parts derivatives are all continuous, how many of them can be distinct or that would be then, um, that would be m plus one.

77E. 78E. 79E. (a) How many n th-order partial derivatives does a function of two variables have? (b) If these partial derivatives are all continuous, how many of them can be distinct? Step-by-step solution. Step 1 of 3. (A) For a function f of two variables the second order partial derivatives are.

... solutions and your answer to the following textbook question: If these partial derivatives are all continuous, how many of them can be distinct?.

18.05.2018 · Show activity on this post. The theorem says that for f to be differentiable, partial derivatives of f exist and are continuous. For example, let f ( x, y) = x 2 + 2 x y + y 2. Let ( a, b) ∈ R 2. Then, I know that partial derivatives exist and f x ( a, b) = 2 a + b, and f y ( a, b) = a + 2 b. In order to test the continuity,

(b) If these partial derivatives are all continuous, how many of them can be distinct? (c) Answer the question in part (a) for a function of three variables ...

How about n variables? If these partial derivatives are all continuous, how many of them can be distinct? Solution. Counting mixed partials as ...