Second derivative of chained function

[lriuui0x0]

Let's say we have a function $M(f(x))$ where $M: \mathbb{R}^2 \to \mathbb{R}^2$ is a multivariable linear function, and $f: \mathbb{R} \to \mathbb{R}^2$ is a single variable function. Now I'm getting confused with evaluating the following second derivative of the function:

$$
[M(f(x))]'' = [M'(f(x)) \circ f'(x)]'
$$

How do we continue to evaluate the second derivative? This may be a very basic question, but I'm trying to get clear on the dimension of functions.

 

[PeroK]

Let's say we have a function $M(f(x))$ where $M: \mathbb{R}^2 \to \mathbb{R}^2$ is a multivariable linear function, and $f: \mathbb{R} \to \mathbb{R}$ is a single variable function. Now I'm getting confused with evaluating the following second derivative of the function:

$$
[M(f(x))]'' = [M'(f(x)) \circ f'(x)]'
$$

How do we continue to evaluate the second derivative? This may be a very basic question, but I'm trying to get clear on the dimension of functions.

This doesn't make sense. If $M$ is a function on $\mathbb R^2$, then it needs two argumemts. You've given it only one, namely $f(x)$.

 

[PeroK]

Let's say we have a function $M(f(x))$ where $M: \mathbb{R}^2 \to \mathbb{R}^2$ is a multivariable linear function, and $f: \mathbb{R} \to \mathbb{R}$ is a single variable function. Now I'm getting confused with evaluating the following second derivative of the function:

$$
[M(f(x))]'' = [M'(f(x)) \circ f'(x)]'
$$

How do we continue to evaluate the second derivative? This may be a very basic question, but I'm trying to get clear on the dimension of functions.

Assuming $M: \mathbb{R} \to \mathbb{R}$, and we want the second derivative of the function $g(x) = M(f(x))$, then the first derivative is: $$g'(x) = M'(f(x))f'(x)$$ Note that this is not what you've written.

The second derivative is then evaluated using the product rule and the chain rule again.

 

[lriuui0x0]

This doesn't make sense. If $M$ is a function on $\mathbb R^2$, then it needs two argumemts. You've given it only one, namely $f(x)$.

Sorry I made a mistake in my question, it should be $f: \mathbb{R} \to \mathbb{R}^2$ instead.

 

[PeroK]

Sorry I made a mistake in my question, it should be $f: \mathbb{R} \to \mathbb{R}^2$ instead.

You have partial derivatives, not ordinary derivatives, for $M(y, z)$. Where $f(x) = (y(x), z(x))$.

 

[lriuui0x0]

You have partial derivatives, not ordinary derivatives, for $M(y, z)$. Where $f(x) = (y(x), z(x))$.

I'm thinking about the total derivatives instead, $M'$ is a map $\mathbb{R}^2 \to \mathbb{R}^2$ at every $\mathbb{R}^2$ point. Therefore $M'(f(x)) : \mathbb{R}^2 \to \mathbb{R}^2$ and it can be composed with $f'(x): \mathbb{R} \to \mathbb{R}^2$.

 

[pasmith]

Let's say we have a function $M(f(x))$ where $M: \mathbb{R}^2 \to \mathbb{R}^2$ is a multivariable linear function, and $f: \mathbb{R} \to \mathbb{R}^2$ is a single variable function. Now I'm getting confused with evaluating the following second derivative of the function:

$$
[M(f(x))]'' = [M'(f(x)) \circ f'(x)]'
$$

How do we continue to evaluate the second derivative? This may be a very basic question, but I'm trying to get clear on the dimension of functions.

$\circ$ is function composition; the inner product is $\cdot$ (\cdot)

The derivative of $M$ is a 2x2 matrix $J_{ij} = \partial_jM_i$; the second derivative is a 2x2x2 tensor $K_{ijk} = \partial_j\partial_k M_i$ where $\delta_i$ denotes differentiation with respect to the $i$th argument. It is difficult to write expressions involving these without using suffix notation:
$$\begin{split}<br /> (M \circ f)_i'(t) &= (J_{ij} \circ f)(t)f_j'(t) \\<br /> (M \circ f)_i''(t) &= (J_{ij} \circ f)(t)f_j''(t) + f_j'(t)f_k'(t)(K_{ijk} \circ f)(t)<br /> \end{split}$$ Now the first derivative and the first term of the second derivative can be written as matrix products of $(J \circ f)(t))$ and $f'(t)$ or $f''(t)$ respectively; it's much harder to do that with the second term of the second derivative, but $((K \circ f)(t) \cdot f'(t)) \cdot f'(t)$ is an attempt, on the understanding that dotting implies contraction over the last index of the first factor and the first index of the second factor.

 

[PeroK]

I'm thinking about the total derivatives instead, $M'$ is a map $\mathbb{R}^2 \to \mathbb{R}^2$ at every $\mathbb{R}^2$ point. Therefore $M'(f(x)) : \mathbb{R}^2 \to \mathbb{R}^2$ and it can be composed with $f'(x): \mathbb{R} \to \mathbb{R}^2$.

You've lost me. I think you need to specify carefully (without typos) what functions you are dealing with and what derivatives you want to calculate. $M'(x)$ is generally only used for a simple function of one variable.

 

[lriuui0x0]

$\circ$ is function composition; the inner product is $\cdot$ (\cdot)

The derivative of $M$ is a 2x2 matrix $J_{ij} = \partial_jM_i$; the second derivative is a 2x2x2 tensor $K_{ijk} = \partial_j\partial_k M_i$ where $\delta_i$ denotes differentiation with respect to the $i$th argument. It is difficult to write expressions involving these without using suffix notation:
$$\begin{split}<br /> (M \circ f)_i'(t) &= (J_{ij} \circ f)(t)f_j'(t) \\<br /> (M \circ f)_i''(t) &= (J_{ij} \circ f)(t)f_j''(t) + f_j'(t)f_k'(t)(K_{ijk} \circ f)(t)<br /> \end{split}$$ Now the first derivative and the first term of the second derivative can be written as matrix products of $(J \circ f)(t))$ and $f'(t)$ or $f''(t)$ respectively; it's much harder to do that with the second term of the second derivative, but $((K \circ f)(t) \cdot f'(t)) \cdot f'(t)$ is an attempt, on the understanding that dotting implies contraction over the last index of the first factor and the first index of the second factor.

Is there some reference of chain rule using index notation? Also what do you mean by $J_{ij} \circ f$, isn't $J_{ij}$ a single number?

 

[lriuui0x0]

You've lost me. I think you need to specify carefully (without typos) what functions you are dealing with and what derivatives you want to calculate. $M'(x)$ is generally only used for a simple function of one variable.

Sorry about that! I guess I'm just randomly coming up with the notation...

 

[S.G. Janssens]

where $M: \mathbb{R}^2 \to \mathbb{R}^2$ is a multivariable linear function

Do you mean that $M$ is linear as a function of its argument in $\mathbb{R}^2$? If yes, then its role in computing the derivative of $M \circ f : \mathbb{R} \to \mathbb{R}^2$ is trivial and you may want to forget about it for the purpose of the question.

 

[lriuui0x0]

Do you mean that $M$ is linear as a function of its argument in $\mathbb{R}^2$? If yes, then its role in computing the derivative of $M \circ f : \mathbb{R} \to \mathbb{R}^2$ is trivial and you may want to forget about it for the purpose of the question.

I'd like to know for both $M$ is linear or not. When $M$ is linear, it's basically a matrix. Why being linear makes the question trivial?

 

[S.G. Janssens]

I'd like to know for both $M$ is linear or not. Why being linear makes the question trivial?

Because then the answer is
$$
(M \circ f)^{(r)}(x) = M(f^{(r)}(x)) \qquad \forall\,r\in \mathbb{Z}_+,\,\forall\,x\in \mathbb{R},
$$
where the superscript indicates the $r$th derivative with respect to the scalar variable $x$. I would also interpret $M$ as a matrix and write the RHS as $M f^{(r)}(x)$. ($M$ just transforms the $r$th derivative of the curve in $\mathbb{R}^2$ parameterized by $f$.)

 

[lriuui0x0]

Because then the answer is
$$
(M \circ f)^{(r)}(x) = M(f^{(r)}(x)) \qquad \forall\,r\in \mathbb{Z}_+,\,\forall\,x\in \mathbb{R},
$$
where the superscript indicates the $r$th derivative with respect to the scalar variable $x$. I would also interpret $M$ as a matrix and write the RHS as $M f^{(r)}(x)$. ($M$ just transforms the $r$th derivative of the curve in $\mathbb{R}^2$ parameterized by $f$.)

Oh is this just the analogy of $(sf)'$ where $s$ is a scalar. Here we just replace the scalar multiplication with a matrix multiplication but it still behaves like a scalar.

 

[S.G. Janssens]

Oh is this just the analogy of $(sf)'$ where $s$ is a scalar. Here we just replace the scalar multiplication with a matrix multiplication but it still behaves like a scalar.

Exactly.