smileycellist2

2021-03-07

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.

casincal

Step 1
Mixed Derivative theorem:" If the function f(x,y) and its partial derivatives ${f}_{x},{f}_{y},{f}_{xy}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{f}_{yx}$ are all defined in any open interval (a,b) and all are continues in the interval, then ${f}_{xy}\left(a,b\right)={f}_{yx}\left(a,b\right)$".
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 $f\left(x,y\right)={x}^{3}{y}^{3}+{x}^{2}y+{y}^{2}x$.
Find the first order partial derivatives as follows.
$\frac{\partial }{\partial x}\left({x}^{3}{y}^{3}+{x}^{2}y+{y}^{2}x\right)=\frac{\partial }{\partial x}\left({x}^{3}{y}^{3}\right)+\frac{\partial }{\partial x}\left({x}^{2}y\right)+\frac{\partial }{\partial x}\left({y}^{2}x\right)$
$=3{y}^{3}{x}^{2}+2yx+{y}^{2}$
$\frac{\partial }{\partial y}\left({x}^{3}{y}^{3}+{x}^{2}y+{y}^{2}x\right)=\frac{\partial }{\partial y}\left({x}^{3}{y}^{3}\right)+\frac{\partial }{\partial y}\left({x}^{2}y\right)+\frac{\partial }{\partial y}\left({y}^{2}x\right)$
$=3{x}^{3}{y}^{2}+{x}^{2}+2xy$
Step 3
Now, find the mixed partial derivatives as,
$\frac{{\partial }^{2}}{\partial y\partial x}\left({x}^{3}{y}^{3}+{x}^{2}y+{y}^{2}x\right)=\frac{\partial }{\partial x}\left(3{x}^{3}{y}^{2}+{x}^{2}+2xy\right)$
$=9{y}^{2}{x}^{2}+2x+2y$
$\frac{{\partial }^{2}}{\partial x\partial y}\left({x}^{3}{y}^{3}+{x}^{2}y+{y}^{2}x\right)=\frac{\partial }{\partial y}\left(3{y}^{3}{x}^{2}+2yx+{y}^{2}\right)$
$=9{x}^{2}{y}^{2}+2x+2y$
That is, $\frac{{\partial }^{2}f}{\partial y\partial x}=\frac{{\partial }^{2}f}{\partial x\partial y}$.

alenahelenash

Mathematically, the Mixed Derivative Theorem can be expressed as follows:
$\frac{{\partial }^{2}f}{\partial x\partial y}=\frac{{\partial }^{2}f}{\partial y\partial x}$
This theorem allows us to simplify the process of calculating partial derivatives of second and higher orders by reducing the number of derivatives we need to evaluate.
Let's consider an example to illustrate this concept. Suppose we have a function $f\left(x,y\right)={x}^{3}{y}^{2}+2xy+3$. We want to find the mixed second-order partial derivatives $\frac{{\partial }^{2}f}{\partial x\partial y}$ and $\frac{{\partial }^{2}f}{\partial y\partial x}$.
To calculate $\frac{{\partial }^{2}f}{\partial x\partial y}$, we differentiate $f$ with respect to $x$ first, and then differentiate the result with respect to $y$. Applying the Chain Rule, we obtain:
$\frac{{\partial }^{2}f}{\partial x\partial y}=\frac{\partial }{\partial y}\left(\frac{\partial f}{\partial x}\right)=\frac{\partial }{\partial y}\left(3{x}^{2}{y}^{2}+2y\right)=6xy$
To calculate $\frac{{\partial }^{2}f}{\partial y\partial x}$, we differentiate $f$ with respect to $y$ first, and then differentiate the result with respect to $x$. Again using the Chain Rule, we get:
$\frac{{\partial }^{2}f}{\partial y\partial x}=\frac{\partial }{\partial x}\left(\frac{\partial f}{\partial y}\right)=\frac{\partial }{\partial x}\left(2x+2x{y}^{2}\right)=2+2{y}^{2}$
As we can see, the mixed second-order partial derivatives $\frac{{\partial }^{2}f}{\partial x\partial y}$ and $\frac{{\partial }^{2}f}{\partial y\partial x}$ are equal to $6xy$ and $2+2{y}^{2}$ respectively. This result confirms the validity of the Mixed Derivative Theorem.
By using the Mixed Derivative Theorem, we can simplify the calculation of mixed partial derivatives by choosing the order of differentiation that is most convenient. This theorem is especially useful when dealing with complex functions involving multiple variables, as it allows us to interchange the order of differentiation and obtain the same result.

user_27qwe

The Mixed Derivative Theorem states that for a function $f\left(x,y\right)$ with continuous second-order partial derivatives, the order of differentiation with respect to two different variables does not matter. In other words, the mixed partial derivatives are equal:
$\frac{{\partial }^{2}f}{\partial x\partial y}=\frac{{\partial }^{2}f}{\partial y\partial x}$
This theorem is helpful in calculating partial derivatives of second and higher orders because it allows us to simplify the process. Instead of calculating all the mixed partial derivatives separately, we only need to compute one of them and then apply the theorem to obtain the others.
For example, consider the function $f\left(x,y\right)={x}^{3}{y}^{2}+2xy$. To find the mixed second-order partial derivatives, we can first calculate $\frac{{\partial }^{2}f}{\partial x\partial y}$:
$\frac{{\partial }^{2}f}{\partial x\partial y}=\frac{\partial }{\partial x}\left(\frac{\partial f}{\partial y}\right)=\frac{\partial }{\partial x}\left(3{x}^{2}{y}^{2}+2y\right)=6x{y}^{2}$
Then, by applying the Mixed Derivative Theorem, we obtain $\frac{{\partial }^{2}f}{\partial y\partial x}$:
$\frac{{\partial }^{2}f}{\partial y\partial x}=\frac{\partial }{\partial y}\left(\frac{\partial f}{\partial x}\right)=\frac{\partial }{\partial y}\left(3{x}^{2}{y}^{2}+2x\right)=6xy$
Therefore, $\frac{{\partial }^{2}f}{\partial x\partial y}=\frac{{\partial }^{2}f}{\partial y\partial x}=6x{y}^{2}=6xy$. The Mixed Derivative Theorem allows us to save time and effort by avoiding redundant calculations.

karton

Result:
$\frac{{\partial }^{2}f}{\partial x\partial y}=6x-4y$ and $\frac{{\partial }^{2}f}{\partial y\partial x}=6x-4y$
Solution:
$\frac{{\partial }^{2}f}{\partial x\partial y}=\frac{{\partial }^{2}f}{\partial y\partial x}$
This theorem is useful in calculating partial derivatives of second and higher orders because it allows us to interchange the order of differentiation without affecting the result.
To illustrate this, let's consider an example. Suppose we have the function $f\left(x,y\right)=3{x}^{2}y-2x{y}^{2}$. We can find the second-order mixed partial derivatives using the Mixed Derivative Theorem.
First, we calculate the first-order partial derivatives:
$\frac{\partial f}{\partial x}=6xy-2{y}^{2}$
$\frac{\partial f}{\partial y}=3{x}^{2}-4xy$
Next, we differentiate these first-order derivatives with respect to the other variable:
$\frac{{\partial }^{2}f}{\partial x\partial y}=\frac{\partial }{\partial y}\left(6xy-2{y}^{2}\right)=6x-4y$
$\frac{{\partial }^{2}f}{\partial y\partial x}=\frac{\partial }{\partial x}\left(3{x}^{2}-4xy\right)=6x-4y$
As we can see, both mixed second-order partial derivatives are equal: $\frac{{\partial }^{2}f}{\partial x\partial y}=6x-4y$ and $\frac{{\partial }^{2}f}{\partial y\partial x}=6x-4y$.

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