# Second derivative composite function

1. Sep 28, 2012

### 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

2. Sep 28, 2012

### haruspex

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

3. Sep 28, 2012

### pasmith

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.

4. Sep 29, 2012

### aliekor

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.

5. Oct 17, 2012

### pasmith

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

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

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

6. Oct 17, 2012

### Muphrid

$$(a * \nabla_t) f = [(\{a * \nabla_t \} g) * \nabla_g] f(g)$$
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,
$$\nabla_t^2 f = [(\nabla_t g) \cdot \nabla_g]^2 f(g)$$