# What is the Mixed Derivative Theorem for mixed second-order partial derivatives? How can it help in calculating partial derivatives of second and higher orders? Give examples. Question
Derivatives What is the Mixed Derivative Theorem for mixed second-order partial derivatives? How can it help in calculating partial derivatives of second and higher orders? Give examples. 2021-03-08
Step 1
Mixed Derivative theorem:" If the function f(x,y) and its partial derivatives $$\displaystyle{f}_{{x}},{f}_{{y}},{f}_{{{x}{y}}}{\quad\text{and}\quad}{f}_{{{y}{x}}}$$ are all defined in any open interval (a,b) and all are continues in the interval, then $$\displaystyle{{f}_{{{x}{y}}}{\left({a},{b}\right)}}={{f}_{{{y}{x}}}{\left({a},{b}\right)}}$$".
That is, mixed derivative theorem says that the mixed partial derivatives are equal.
Thus, there is no need of calculating all the mixed partial derivatives. Only one case is enough.
Step 2
For example consider the function $$\displaystyle{f{{\left({x},{y}\right)}}}={x}^{{3}}{y}^{{3}}+{x}^{{2}}{y}+{y}^{{2}}{x}$$.
Find the first order partial derivatives as follows.
$$\displaystyle\frac{\partial}{{\partial{x}}}{\left({x}^{{3}}{y}^{{3}}+{x}^{{2}}{y}+{y}^{{2}}{x}\right)}=\frac{\partial}{{\partial{x}}}{\left({x}^{{3}}{y}^{{3}}\right)}+\frac{\partial}{{\partial{x}}}{\left({x}^{{2}}{y}\right)}+\frac{\partial}{{\partial{x}}}{\left({y}^{{2}}{x}\right)}$$
$$\displaystyle={3}{y}^{{3}}{x}^{{2}}+{2}{y}{x}+{y}^{{2}}$$
$$\displaystyle\frac{\partial}{{\partial{y}}}{\left({x}^{{3}}{y}^{{3}}+{x}^{{2}}{y}+{y}^{{2}}{x}\right)}=\frac{\partial}{{\partial{y}}}{\left({x}^{{3}}{y}^{{3}}\right)}+\frac{\partial}{{\partial{y}}}{\left({x}^{{2}}{y}\right)}+\frac{\partial}{{\partial{y}}}{\left({y}^{{2}}{x}\right)}$$
$$\displaystyle={3}{x}^{{3}}{y}^{{2}}+{x}^{{2}}+{2}{x}{y}$$
Step 3
Now, find the mixed partial derivatives as,
$$\displaystyle\frac{\partial^{{2}}}{{\partial{y}\partial{x}}}{\left({x}^{{3}}{y}^{{3}}+{x}^{{2}}{y}+{y}^{{2}}{x}\right)}=\frac{\partial}{{\partial{x}}}{\left({3}{x}^{{3}}{y}^{{2}}+{x}^{{2}}+{2}{x}{y}\right)}$$
$$\displaystyle={9}{y}^{{2}}{x}^{{2}}+{2}{x}+{2}{y}$$
$$\displaystyle\frac{\partial^{{2}}}{{\partial{x}\partial{y}}}{\left({x}^{{3}}{y}^{{3}}+{x}^{{2}}{y}+{y}^{{2}}{x}\right)}=\frac{\partial}{{\partial{y}}}{\left({3}{y}^{{3}}{x}^{{2}}+{2}{y}{x}+{y}^{{2}}\right)}$$
$$\displaystyle={9}{x}^{{2}}{y}^{{2}}+{2}{x}+{2}{y}$$
That is, $$\displaystyle\frac{{\partial^{{2}}{f}}}{{\partial{y}\partial{x}}}=\frac{{\partial^{{2}}{f}}}{{\partial{x}\partial{y}}}$$.

### Relevant Questions Mixed Partial Derivatives If f is a function of a and y such that $$\displaystyle{f}_{{{x}{y}}}{\quad\text{and}\quad}{f}_{{{y}{x}}}$$ are continuous, what is the relationship between the mixed partial derivatives? Consider a function $$\displaystyle{z}={x}{y}+{x}{\left({y}^{{2}}+{1}\right)}$$.Find first order partial derivatives, total differential, and total derivative with respect to x. Calculating derivatives Find the derivative of the following functions.
$$\displaystyle{y}={{\cos}^{{2}}{x}}$$ How can evaluate partial derivatives, gradients, and directional derivatives at a point? Find all first and the second partial derivatives.
$$\displaystyle{f{{\left({x},{y}\right)}}}={2}{x}^{{5}}{y}^{{2}}+{x}^{{2}}{y}$$ Find the four second partial derivatives of $$\displaystyle{f{{\left({x},{y}\right)}}}={x}^{{2}}{y}^{{3}}$$. For the following sets of variables, find all the relevant second derivatives. In all cases, first find general expressions for the second derivatives and then substitute variables at the last step. $$\displaystyle{f{{\left({x},{y}\right)}}}={x}^{{2}}{y}-{x}{y}^{{2}},{w}{h}{e}{r}{e}{x}={s}{t}{\quad\text{and}\quad}{y}=\frac{{s}}{{t}}$$ Higher-Order Derivatives
$$\displaystyle{f{{\left({x}\right)}}}={2}{x}^{{4}}-{3}{x}^{{3}}+{2}{x}^{{2}}+{x}+{4}$$
find $$\displaystyle{{f}^{{10}}{\left({x}\right)}}=?$$ Define Higher order Derivatives ? The table below shows the number of people for three different race groups who were shot by police that were either armed or unarmed. These values are very close to the exact numbers. They have been changed slightly for each student to get a unique problem.
Suspect was Armed:
Black - 543
White - 1176
Hispanic - 378
Total - 2097
Suspect was unarmed:
Black - 60
White - 67
Hispanic - 38
Total - 165
Total:
Black - 603
White - 1243
Hispanic - 416
Total - 2262
Give your answer as a decimal to at least three decimal places.
a) What percent are Black?
b) What percent are Unarmed?
c) In order for two variables to be Independent of each other, the P $$(A and B) = P(A) \cdot P(B) P(A and B) = P(A) \cdot P(B).$$
This just means that the percentage of times that both things happen equals the individual percentages multiplied together (Only if they are Independent of each other).
Therefore, if a person's race is independent of whether they were killed being unarmed then the percentage of black people that are killed while being unarmed should equal the percentage of blacks times the percentage of Unarmed. Let's check this. Multiply your answer to part a (percentage of blacks) by your answer to part b (percentage of unarmed).
Remember, the previous answer is only correct if the variables are Independent.
d) Now let's get the real percent that are Black and Unarmed by using the table?
If answer c is "significantly different" than answer d, then that means that there could be a different percentage of unarmed people being shot based on race. We will check this out later in the course.
Let's compare the percentage of unarmed shot for each race.
e) What percent are White and Unarmed?
f) What percent are Hispanic and Unarmed?
If you compare answers d, e and f it shows the highest percentage of unarmed people being shot is most likely white.
Why is that?
This is because there are more white people in the United States than any other race and therefore there are likely to be more white people in the table. Since there are more white people in the table, there most likely would be more white and unarmed people shot by police than any other race. This pulls the percentage of white and unarmed up. In addition, there most likely would be more white and armed shot by police. All the percentages for white people would be higher, because there are more white people. For example, the table contains very few Hispanic people, and the percentage of people in the table that were Hispanic and unarmed is the lowest percentage.
Think of it this way. If you went to a college that was 90% female and 10% male, then females would most likely have the highest percentage of A grades. They would also most likely have the highest percentage of B, C, D and F grades
The correct way to compare is "conditional probability". Conditional probability is getting the probability of something happening, given we are dealing with just the people in a particular group.
g) What percent of blacks shot and killed by police were unarmed?
h) What percent of whites shot and killed by police were unarmed?
i) What percent of Hispanics shot and killed by police were unarmed?
You can see by the answers to part g and h, that the percentage of blacks that were unarmed and killed by police is approximately twice that of whites that were unarmed and killed by police.
j) Why do you believe this is happening?
Do a search on the internet for reasons why blacks are more likely to be killed by police. Read a few articles on the topic. Write your response using the articles as references. Give the websites used in your response. Your answer should be several sentences long with at least one website listed. This part of this problem will be graded after the due date.
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