# Second derivative composite function

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

haruspex
Homework Helper
Gold Member
2020 Award
Looks ok to me. Try it with some simple examples.

pasmith
Homework Helper
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.

pasmith
Homework Helper
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

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

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