# Partial derivative of composition

• A
• SchroedingersLion
In summary, the conversation revolves around the interpretation of the subscript in the term ##\nabla_x C(x_0, y(x_0))##, with one person suggesting it represents taking the partial derivative with respect to the variable ##x## and then inserting specific values for ##x## and ##y##, while the other suggests it represents the gradient of the function ##C## with respect to the vector variable ##\vec x## and then plugging in specific values for ##\vec x## and ##\vec y##. The correct interpretation depends on the context of the problem at hand.
SchroedingersLion
Hi guys,

suppose we have a function ##C(x, y)## into the real numbers. Suppose also that ##y=y(x)##, i.e. ##y## is a function of ##x##.

Now in my script, I have a term ##\nabla_x C(x_0, y(x_0)) ##. From my point of view, this means that you take the partial derivative of ##C(x,y)## with respect to x and then insert ##y(x_0)## for ##y##, and ##x_0## for ##x##.
I would not consider the x-derivative of ##y(x)##, since then the subscript at the nabla operator wouldn't make any sense as this would simply be ##\nabla C(x_0, y(x_0))##.

Is this line of thinking correct?

Best

SL.

SchroedingersLion said:
From my point of view, this means that you take the partial derivative of ##C(x,y)## with respect to x and then insert ##y(x_0)## for ##y##, and ##x_0## for ##x##.
I agree.
I would not consider the x-derivative of ##y(x)##, since then the subscript at the nabla operator wouldn't make any sense as this would simply be ##\nabla C(x_0, y(x_0))##.
I would interpret the last expression as vector with the two partial derivatives as components.

SchroedingersLion
SchroedingersLion said:
Hi guys,

suppose we have a function ##C(x, y)## into the real numbers. Suppose also that ##y=y(x)##, i.e. ##y## is a function of ##x##.

In this case you must be careful about what function you are talking about and how you are differentiating. Technically, you have now defined a new function of a single variable:
$$g(x) = C(x, y(x))$$
And, the derivative of ##g## is given by:
$$g'(x) = \frac{\partial C}{\partial x}+ \frac{\partial C}{\partial y}y'(x)$$
It would make no sense, of course, to take the gradient of ##g##. And:
$$g'(x_0) = \frac{\partial C}{\partial x}(x_0, y(x_0)) + \frac{\partial C}{\partial y}(x_0, y(x_0))y'(x_0)$$

However, you have another function which is the gradient of ##C##:
$$\nabla C = \frac{\partial C}{\partial x} \hat x + \frac{\partial C}{\partial y} \hat y$$
Note that this function is also a function of two variables. And, you can now define another function by:
$$h(x) = \nabla C(x, y(x)) = \frac{\partial C}{\partial x}(x, y(x)) \hat x + \frac{\partial C}{\partial y}(x, y(x)) \hat y$$
And, of course:
$$h(x_0) = \nabla C(x_0, y(x_0)) = \frac{\partial C}{\partial x}(x_0, y(x_0)) \hat x + \frac{\partial C}{\partial y}(x_0, y(x_0)) \hat y$$
In the context of your question, therefore, you need to decide which function you are dealing with: ##g## or ##h##.

SchroedingersLion
Thanks for the responses!

The two of you seem to disagree about my own interpretation what to do.

@PeroK
I should have noted that ##x## and ##y## are each in ##R^d## so the partial derivatives should be gradients.
From your explanations, I still don't see what the subscript then means in ##\nabla_x##. If ##C(x, y(x))## implies the redefinition of ##C(x,y)## as a function ##g(x)##, then the subscript of the gradient would be useless, right?
Therefore, my original idea:
SchroedingersLion said:
From my point of view, this means that you take the partial derivative of ##C(x,y)## with respect to x and then insert ##y(x_0)## for ##y##, and ##x_0## for ##x##.

SchroedingersLion said:
Thanks for the responses!

The two of you seem to disagree about my own interpretation what to do.

@PeroK
I should have noted that ##x## and ##y## are each in ##R^d## so the partial derivatives should be gradients.
From your explanations, I still don't see what the subscript then means in ##\nabla_x##. If ##C(x, y(x))## implies the redefinition of ##C(x,y)## as a function ##g(x)##, then the subscript of the gradient would be useless, right?
Therefore, my original idea:

Okay, so you actually have a function of two vector variables: ##C(\vec x, \vec y) = C(x_1, x_2, x_3, y_1, y_2, y_3)##.

There's the same issue. Are you first reducing ##C## to a function of a single vector ##\vec x##, using ##\vec y = \vec y(\vec x)##? Or, are you taking the gradient of ##C## with respect to ##\vec x## and then plugging in ##\vec x_0, \vec y_0##?

Which one of these it is depends on the context of what you are doing.

PeroK said:
Okay, so you actually have a function of two vector variables: ##C(\vec x, \vec y) = C(x_1, x_2, x_3, y_1, y_2, y_3)##.

There's the same issue. Are you first reducing ##C## to a function of a single vector ##\vec x##, using ##\vec y = \vec y(\vec x)##? Or, are you taking the gradient of ##C## with respect to ##\vec x## and then plugging in ##\vec x_0, \vec y_0##?

Which one of these it is depends on the context of what you are doing.

That's essentially what I don't know. I had hoped there was some clear convention =(

SchroedingersLion said:
That's essentially what I don't know. I had hoped there was some clear convention =(

I don't think it's a matter of convention. It's a matter of the author being unambiguous. You should be able to work out which one it is from what follows.

## 1. What is a partial derivative of composition?

The partial derivative of composition is a mathematical concept that involves taking the derivative of a function that is composed of two or more functions. It is used to measure the rate of change of a dependent variable with respect to one of its independent variables while holding the other variables constant.

## 2. How is the partial derivative of composition calculated?

The partial derivative of composition is calculated by first finding the derivative of the outer function, and then multiplying it by the derivative of the inner function. This process is repeated for each variable in the composition.

## 3. What is the purpose of calculating the partial derivative of composition?

The partial derivative of composition is used to analyze the relationship between multiple variables in a function. It can help determine the sensitivity of the dependent variable to changes in the independent variables and identify critical points or extrema in the function.

## 4. Can the partial derivative of composition be applied to any type of function?

Yes, the partial derivative of composition can be applied to any type of function, including polynomial, exponential, logarithmic, and trigonometric functions. It is a fundamental tool in multivariable calculus and is used in many fields of science and engineering.

## 5. Are there any real-world applications of the partial derivative of composition?

Yes, the partial derivative of composition has many real-world applications in fields such as physics, economics, and engineering. For example, it can be used to analyze the relationship between supply and demand in economics, or to optimize the design of a structure in engineering.

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