Second derivative composite function

• aliekor
In summary, the second time derivative of f can be found by taking the second derivative of f with respect to g and multiplying it by g'
aliekor
Hi guys, I have this function

f(g(t)) and I have to find the second time derivative of f, is it correct the following solution?:

f''=∂f/∂g*g'=∇f*g'
f ''=∇^2f*|g'|^2+∇f*g''

where ∇^2 is the laplacian function

Looks ok to me. Try it with some simple examples.

Your (standard) notational abuse has confused you.

The chain rule for a vector-valued function (f) of a vector-valued function (g) of a scalar (t) is:
$$\frac{d\mathbf{f}(\mathbf{g})}{dt} = (\nabla \mathbf{f}) \cdot \frac{d\mathbf{g}}{dt}$$
ie
$$\frac{df_i(\mathbf{g})}{dt}(t) = \sum_j \frac{\partial f_i}{\partial x_j}(\mathbf{g}(t)) \frac{dg_j}{dt}(t).$$

Differentiating with respect to t and using the product rule gives:
$$\frac{d^2\mathbf{f}(\mathbf{g})}{dt^2} = \left(\frac{d}{dt}(\nabla \mathbf{f})\right) \cdot \frac{d\mathbf{g}}{dt} + (\nabla \mathbf{f}) \cdot \frac{d^2\mathbf{g}}{dt^2}$$
Using the chain rule on the first term gives
$$\frac{d}{dt}(\nabla \mathbf{f}) = (\nabla \nabla \mathbf{f})\cdot \frac{d\mathbf{g}}{dt}$$
and so
$$\frac{d^2\mathbf{f}(\mathbf{g})}{dt^2} = \left(\left((\nabla \nabla \mathbf{f}) \cdot \frac{d\mathbf{g}}{dt}\right) \cdot \frac{d\mathbf{g}}{dt}\right) + (\nabla \mathbf{f}) \cdot \frac{d^2\mathbf{g}}{dt^2}$$
ie
$$\frac{d^2f_i(\mathbf{g})}{dt^2} = \sum_j \sum_k \frac{dg_j}{dt} \frac {dg_k}{dt} \frac{\partial^2 f_i}{\partial x_j \partial x_k} + \sum_j \frac{d^2g_j}{dt^2} \frac{\partial f_i}{\partial x_j}.$$

You appear to have the first term as
$$\sum_j \sum_k \left(\frac {dg_j}{dt}\right)^2 \frac{\partial^2 f_i}{\partial x_k^2}$$
which is incorrect.

Thank you pasmith, actually I didn't get comething with the dimensions:
if I have g as a vector 2x1
$$\nabla \nabla \mathbf{f}$$
will be a 2x2 matrix. In the calculus of :
$$\left(\left((\nabla \nabla \mathbf{f}) \cdot \frac{d\mathbf{g}}{dt}\right) \cdot \frac{d\mathbf{g}}{dt}\right)$$
I have a matrix 2x2 times a vector 2X1 which gives a vector 2x1 and this vector is multiplied by a vector 2x1 but the dimension are wrong.

aliekor said:
Thank you pasmith, actually I didn't get comething with the dimensions:
if I have g as a vector 2x1
$$\nabla \nabla \mathbf{f}$$
will be a 2x2 matrix.

Actually it's a 2x2x2 (ie, rank 3) tensor:
$$(\nabla\nabla \mathbf{f})_{ijk} = \frac{\partial^2 f_i}{\partial x_j \partial x_k}$$

In the calculus of :
$$\left(\left((\nabla \nabla \mathbf{f}) \cdot \frac{d\mathbf{g}}{dt}\right) \cdot \frac{d\mathbf{g}}{dt}\right)$$
I have a matrix 2x2 times a vector 2X1 which gives a vector 2x1 and this vector is multiplied by a vector 2x1 but the dimension are wrong.

This is a rank 3 tensor dotted with two vectors (rank 1 tensors), which is equivalent to a
rank 5 tensor contracted over two pairs of indices, yielding a rank 1 tensor (ie a vector).
In cartesian components, it is

$$\sum_{j}\sum_{k} \frac{\partial^2 f_i}{\partial x_j \partial x_k} \frac{dg_j}{dt} \frac{dg_k}{dt}$$

aliekor said:
Hi guys, I have this function

f(g(t)) and I have to find the second time derivative of f, is it correct the following solution?:

f''=∂f/∂g*g'=∇f*g'
f ''=∇^2f*|g'|^2+∇f*g''

where ∇^2 is the laplacian function

$$(a * \nabla_t) f = [(\{a * \nabla_t \} g) * \nabla_g] f(g)$$
This is the chain rule. The stars just mean a generalized product. This form of the chain rule is valid for ##t## vector or scalar. Here, ##a## must be a scalar, so we'll just cancel it out.

$$\nabla_t f = [(\nabla_t g) \cdot \nabla_g] f(g)$$

Clearly, then,
$$\nabla_t^2 f = [(\nabla_t g) \cdot \nabla_g]^2 f(g)$$

1. What is a second derivative composite function?

A second derivative composite function is a mathematical function that is composed of two or more functions, where the second derivative of the outer function is multiplied by the derivative of the inner function.

2. What is the purpose of finding the second derivative of a composite function?

The second derivative of a composite function allows us to analyze the rate of change of the rate of change of the original function. This can provide insight into the concavity and inflection points of the function, as well as the maximum and minimum values.

3. How do you find the second derivative of a composite function?

To find the second derivative of a composite function, you first find the derivative of the outer function, then multiply it by the derivative of the inner function, and finally take the derivative of the inner function again and multiply it by the second derivative of the outer function.

4. Can you give an example of a second derivative composite function?

One example of a second derivative composite function is f(x) = (x^2 + 3x)^2. The first derivative would be f'(x) = 2(x^2 + 3x)(2x + 3) and the second derivative would be f''(x) = 2(2x + 3)^2 + 2(2)(x^2 + 3x)(2) = 4(2x + 3)(x^2 + 3x + 1).

5. How is the second derivative of a composite function used in real life?

The second derivative of a composite function is used in various fields of science and engineering to analyze and model physical phenomena, such as the acceleration of a moving object or the rate of change of a chemical reaction. It is also used in economics and finance to analyze the curvature and stability of a market trend.

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