Partial derivative of composition

[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.

 

[mfb]

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.

 

[PeroK]

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:

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 $x$.

 

[PeroK]

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.

 

[SchroedingersLion]

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 =(

 

[PeroK]

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.