Pieszkowo3gc4

2023-03-30

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?

coguarsbq2q

Beginner2023-03-31Added 10 answers

The Mixed Derivative Theorem states that for a function $f(x,y)$ with continuous second-order partial derivatives, the mixed partial derivatives $\frac{{\partial}^{2}f}{\partial y\partial x}$ and $\frac{{\partial}^{2}f}{\partial x\partial y}$ are equal.

Mathematically, the Mixed Derivative Theorem can be expressed as:

$\frac{{\partial}^{2}f}{\partial y\partial x}=\frac{{\partial}^{2}f}{\partial x\partial y}$

This theorem implies that the order in which we take the partial derivatives with respect to different variables does not matter, as long as the function has continuous second-order partial derivatives.

The Mixed Derivative Theorem is useful in calculating partial derivatives of second and higher orders because it allows us to simplify calculations and avoid redundancy. Instead of calculating both $\frac{{\partial}^{2}f}{\partial y\partial x}$ and $\frac{{\partial}^{2}f}{\partial x\partial y}$ separately, we only need to calculate one of them, as they are equal.

Here's an example to illustrate the concept:

Consider the function $f(x,y)=3{x}^{2}y+{y}^{3}$.

We can calculate the second-order partial derivatives using the Mixed Derivative Theorem:

First, we find the first-order partial derivatives:

$\frac{\partial f}{\partial x}=6xy$

$\frac{\partial f}{\partial y}=3{x}^{2}+3{y}^{2}$

Next, we calculate the second-order partial derivatives:

$\frac{{\partial}^{2}f}{\partial y\partial x}=\frac{\partial}{\partial y}\left(\frac{\partial f}{\partial x}\right)=\frac{\partial}{\partial y}(6xy)=6x$

$\frac{{\partial}^{2}f}{\partial x\partial y}=\frac{\partial}{\partial x}\left(\frac{\partial f}{\partial y}\right)=\frac{\partial}{\partial x}(3{x}^{2}+3{y}^{2})=6x$

As we can see, the mixed partial derivatives $\frac{{\partial}^{2}f}{\partial y\partial x}$ and $\frac{{\partial}^{2}f}{\partial x\partial y}$ are equal, confirming the validity of the Mixed Derivative Theorem.

Thus, the Mixed Derivative Theorem simplifies the calculation of partial derivatives of second and higher orders by reducing the number of calculations required.

Mathematically, the Mixed Derivative Theorem can be expressed as:

$\frac{{\partial}^{2}f}{\partial y\partial x}=\frac{{\partial}^{2}f}{\partial x\partial y}$

This theorem implies that the order in which we take the partial derivatives with respect to different variables does not matter, as long as the function has continuous second-order partial derivatives.

The Mixed Derivative Theorem is useful in calculating partial derivatives of second and higher orders because it allows us to simplify calculations and avoid redundancy. Instead of calculating both $\frac{{\partial}^{2}f}{\partial y\partial x}$ and $\frac{{\partial}^{2}f}{\partial x\partial y}$ separately, we only need to calculate one of them, as they are equal.

Here's an example to illustrate the concept:

Consider the function $f(x,y)=3{x}^{2}y+{y}^{3}$.

We can calculate the second-order partial derivatives using the Mixed Derivative Theorem:

First, we find the first-order partial derivatives:

$\frac{\partial f}{\partial x}=6xy$

$\frac{\partial f}{\partial y}=3{x}^{2}+3{y}^{2}$

Next, we calculate the second-order partial derivatives:

$\frac{{\partial}^{2}f}{\partial y\partial x}=\frac{\partial}{\partial y}\left(\frac{\partial f}{\partial x}\right)=\frac{\partial}{\partial y}(6xy)=6x$

$\frac{{\partial}^{2}f}{\partial x\partial y}=\frac{\partial}{\partial x}\left(\frac{\partial f}{\partial y}\right)=\frac{\partial}{\partial x}(3{x}^{2}+3{y}^{2})=6x$

As we can see, the mixed partial derivatives $\frac{{\partial}^{2}f}{\partial y\partial x}$ and $\frac{{\partial}^{2}f}{\partial x\partial y}$ are equal, confirming the validity of the Mixed Derivative Theorem.

Thus, the Mixed Derivative Theorem simplifies the calculation of partial derivatives of second and higher orders by reducing the number of calculations required.

How do I find the y-intercept of a parabola?

What are the vertices of $9{x}^{2}+16{y}^{2}=144$?

How to determine the rate of change of a function?

Why are the tangents for 90 and 270 degrees undefined?

How to find the center and radius of the circle ${x}^{2}+{y}^{2}-6x+8y=0$?

What is multiplicative inverse of a number?

How to find the continuity of a function on a closed interval?

How do I find the tangent line of a function?

How to find vertical asymptotes using limits?

How to find the center and radius of the circle ${x}^{2}-12x+{y}^{2}+4y+15=0$?

Let f be a function so that (below). Which must be true?

I. f is continuous at x=2

II. f is differentiable at x=2

III. The derivative of f is continuous at x=2

(A) I (B) II (C) I and II (D) I and III (E) II and IIIHow to find the center and radius of the circle given ${x}^{2}+{y}^{2}+8x-6y=0$?

How to find the center and radius of the circle ${x}^{2}+{y}^{2}+4x-8y+4=0$?

How to identify the center and radius of the circle ${(x+3)}^{2}+{(y-8)}^{2}=16$?

How to evaluate the limit $lim\frac{{e}^{t}}{t}$ as $t\to \infty$?