# Partial Differentiation: second partial derivative

• jellicorse
In summary, z = f(x, y), with x = g(u, v) and y = h(u, v). Here x and y are dependent variables (depending on u and v), and u and v are independent variables. Since u and v are independent (are unrelated to each other), \frac{\partial v}{\partial u} = 0, so your third term vanishes.
jellicorse
I am not quite sure how $$\frac{\partial}{\partial u}\left(\frac{\partial z}{\partial u}\right)$$
$$=\frac{\partial}{\partial u}\left( u \frac{\partial z}{\partial x}+v\frac{\partial z}{\partial y} \right)$$

comes to $$\frac{\partial z}{\partial x} + u\frac{\partial}{\partial u}\left(\frac{\partial z}{\partial x}\right)+v \frac{\partial}{\partial u}\left(\frac{\partial z}{\partial y}\right)$$...

I would have evaluated this to be the same but without the $\frac{\partial z}{\partial x}$ term. i.e.:

$$u\frac{\partial}{\partial u}\left(\frac{\partial z}{\partial x}\right)+v \frac{\partial}{\partial u}\left(\frac{\partial z}{\partial y}\right)$$

I know I am probably overlooking something obvious/simple but I can't see what it is. Could anyone tell me?

jellicorse said:
I am not quite sure how $$\frac{\partial}{\partial u}\left(\frac{\partial z}{\partial u}\right)$$
$$=\frac{\partial}{\partial u}\left( u \frac{\partial z}{\partial x}+v\frac{\partial z}{\partial y} \right)$$
How are z, u, and v related? I can guess, but that really is information you should supply.
jellicorse said:
comes to $$\frac{\partial z}{\partial x} + u\frac{\partial}{\partial u}\left(\frac{\partial z}{\partial x}\right)+v \frac{\partial}{\partial u}\left(\frac{\partial z}{\partial y}\right)$$...

I would have evaluated this to be the same but without the $\frac{\partial z}{\partial x}$ term. i.e.:

$$u\frac{\partial}{\partial u}\left(\frac{\partial z}{\partial x}\right)+v \frac{\partial}{\partial u}\left(\frac{\partial z}{\partial y}\right)$$

I know I am probably overlooking something obvious/simple but I can't see what it is. Could anyone tell me?

Sorry. $z =f (x,y)$, with

$$x=\frac{1}{2}(u^2-v^2)$$ and $y =uv$

Do you understand this part?
$$\frac{\partial z}{\partial u}=\left( u \frac{\partial z}{\partial x}+v\frac{\partial z}{\partial y} \right)$$

Or are you having difficulties with this part?
$$\frac{\partial}{\partial u}( u \frac{\partial z}{\partial x}+v\frac{\partial z}{\partial y} )$$

$$= \frac{\partial z}{\partial x} + u\frac{\partial}{\partial u}\left(\frac{\partial z}{\partial x}\right)+v \frac{\partial}{\partial u} \left( \frac{\partial z}{\partial y} \right)$$

Mark44 said:
Or are you having difficulties with this part?
$$\frac{\partial}{\partial u}( u \frac{\partial z}{\partial x}+v\frac{\partial z}{\partial y} )$$

$$= \frac{\partial z}{\partial x} + u\frac{\partial}{\partial u}\left(\frac{\partial z}{\partial x}\right)+v \frac{\partial}{\partial u} \left( \frac{\partial z}{\partial y} \right)$$

This is the part that I don't understand. The first part is OK...

Mark44 said:
Or are you having difficulties with this part?
$$\frac{\partial}{\partial u}( u \frac{\partial z}{\partial x}+v\frac{\partial z}{\partial y} )$$

$$= \frac{\partial z}{\partial x} + u\frac{\partial}{\partial u}\left(\frac{\partial z}{\partial x}\right)+v \frac{\partial}{\partial u} \left( \frac{\partial z}{\partial y} \right)$$

jellicorse said:
This is the part that I don't understand. The first part is OK...
$$\frac{\partial}{\partial u}( u \frac{\partial z}{\partial x}+v\frac{\partial z}{\partial y} )$$
$$= \frac{\partial}{\partial u}( u \frac{\partial z}{\partial x}) + \frac{\partial}{\partial u}(v\frac{\partial z}{\partial y} )$$

Now use the product rule on each part.

1 person
Thanks a lot, Mark... I am getting closer but have one problem. Having done it several times over, the same thing keeps happening:

Using the product rule $(uv)' = u'v + uv'$ I get one extra term.

I'll type it piece by piece for clarity:

$$\frac{\partial}{\partial u}( u \frac{\partial z}{\partial x}) = (uv)'$$

Gives:

$$\frac{\partial z}{\partial x}+u\frac{\partial}{\partial u}( \frac{\partial z}{\partial x})$$

And

$$\frac{\partial}{\partial u}(v\frac{\partial z}{\partial y} ) = (uv)'$$

Gives:

$$\frac{\partial v}{\partial u}\cdot \frac{\partial z}{\partial y}+v\frac{\partial}{\partial u}(\frac{\partial z}{\partial y} )$$

So, in total I get:

$$\frac{\partial}{\partial u}( u \frac{\partial z}{\partial x})+\frac{\partial}{\partial u}(v\frac{\partial z}{\partial y} )$$

$$=\frac{\partial z}{\partial x}+u\frac{\partial}{\partial u}( \frac{\partial z}{\partial x})+\frac{\partial v}{\partial u}\cdot \frac{\partial z}{\partial y}+v\frac{\partial}{\partial u}(\frac{\partial z}{\partial y} )$$

You have z = f(x, y), where x = g(u, v) and y = h(u, v). Here x and y are dependent variables (depending on u and v), and u and v are independent variables. Since u and v are independent (are unrelated to each other), ## \frac{\partial v}{\partial u} = 0##, so your third term vanishes. By the same reasoning, ## \frac{\partial u}{\partial v}## would also be zero.

Nice job on the LaTeX formatting!

1 person
Oh, I see now. Thanks a lot, it all makes perfect sense now!

## 1. What is partial differentiation?

Partial differentiation is a mathematical technique used to find the rate of change of a multivariate function with respect to one of its variables while holding all other variables constant. It is a tool commonly used in fields such as physics, engineering, and economics to analyze and optimize complex systems.

## 2. What is a second partial derivative?

A second partial derivative is the derivative of a function with respect to two different variables. It measures the rate of change of the rate of change of the function with respect to two different variables, and is denoted by symbols such as fxx or fyy.

## 3. Why is the second partial derivative important?

The second partial derivative is important because it provides valuable information about the shape and behavior of a multivariate function. It can help determine the direction in which the function is increasing or decreasing, the location of critical points, and the curvature of the function's graph.

## 4. How is the second partial derivative calculated?

The second partial derivative is calculated by taking the derivative of the first partial derivative with respect to the other variable. For example, to find fxx, we first find fx (the first partial derivative with respect to x) and then take the derivative of fx with respect to x again.

## 5. What are some real-world applications of the second partial derivative?

The second partial derivative has many applications in fields such as physics, engineering, and economics. For example, it is used to analyze the behavior of electric and magnetic fields in electromagnetism, to optimize the design of structures in engineering, and to determine the optimal production levels in economics.

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