The derivatives of scalars, vectors, and secondorder tensors with respect to secondorder tensors are of considerable use in continuum mechanics. These derivatives are used in the theories of nonlinear elasticity and plasticity, particularly in the design of algorithms for numerical simulations.^{[1]}
The directional derivative provides a systematic way of finding these derivatives.^{[2]}
The definitions of directional derivatives for various situations are given below. It is assumed that the functions are sufficiently smooth that derivatives can be taken.
Let be a real valued function of the vector . Then the derivative of with respect to (or at ) in the direction is the vector defined as
for all vectors .
Properties:
1) If then
2) If then
3) If then
Let be a vector valued function of the vector . Then the derivative of with respect to (or at ) in the direction is the second order tensor defined as
for all vectors .
Let be a real valued function of the second order tensor . Then the derivative of with respect to (or at ) in the direction is the second order tensor defined as
for all second order tensors .
Let be a second order tensor valued function of the second order tensor . Then the derivative of with respect to (or at ) in the direction is the fourth order tensor defined as
for all second order tensors .
The gradient, , of a tensor field in the direction of an arbitrary constant vector is defined as:
The gradient of a tensor field of order n is a tensor field of order n + 1.
If are the basis vectors in a Cartesian coordinate system, with coordinates of points denoted by (x_{1},x_{2},x_{ 3}), then the gradient of the tensor field is given by
Proof 

The vectors and can be written as and . Let . In that case the gradient is given by 
Since the basis vectors do not vary in a Cartesian coordinate system we have the following relations for the gradients of a scalar field φ, a vector field , and a secondorder tensor field .
If are the contravariant basis vectors in a curvilinear coordinate system, with coordinates of points denoted by (ξ^{1},ξ^{2},ξ^{3}), then the gradient of the tensor field is given by (see ^{[3]} for a proof.)
From this definition we have the following relations for the gradients of a scalar field φ, a vector field , and a secondorder tensor field .
where the Christoffel symbol is defined using
In cylindrical coordinates, the gradient is given by
The divergence of a tensor field is defined using the recursive relation
where is an arbitrary constant vector and is a vector field. If is a tensor field of order n > 1 then the divergence of the field is a tensor of order n − 1.
In a Cartesian coordinate system we have the following relations for the divergences of a vector field and a secondorder tensor field .
In curvilinear coordinates, the divergences of a vector field and a secondorder tensor field are
In cylindrical polar coordinates
The curl of an ordern > 1 tensor field is also defined using the recursive relation
where is an arbitrary constant vector and is a vector field.
Consider a vector field and an arbitrary constant vector . In index notation, the cross product is given by
where e_{ijk} is the permutation symbol. Then,
Therefore
For a secondorder tensor
Hence, using the definition of the curl of a firstorder tensor field,
Therefore, we have
The most commonly used identity involving the curl of a tensor field, , is
This identity hold for tensor fields of all orders. For the important case of a secondorder tensor, , this identity implies that
The derivative of the determinant of a second order tensor is given by
In an orthonormal basis, the components of can be written as a matrix . In that case, the right hand side corresponds the cofactors of the matrix.
Proof 

Let be a second order tensor and let . Then, from the definition of the derivative of a scalar valued function of a tensor, we have Recall that we can expand the determinant of a tensor in the form of a characteristic equation in terms of the invariants I_{1},I_{2},I_{ 3} using (note the sign of λ) Using this expansion we can write Recall that the invariant I_{1} is given by Hence, Invoking the arbitrariness of we then have 
The principal invariants of a second order tensor are
The derivatives of these three invariants with respect to are
Proof 

From the derivative of the determinant we know that
For the derivatives of the other two invariants, let us go back to the characteristic equation Using the same approach as for the determinant of a tensor, we can show that Now the left hand side can be expanded as Hence or, Expanding the right hand side and separating terms on the left hand side gives or, If we define I_{0}: = 1 and I_{4}: = 0, we can write the above as Collecting terms containing various powers of λ, we get Then, invoking the arbitrariness of λ, we have This implies that 
Let be the second order identity tensor. Then the derivative of this tensor with respect to a second order tensor is given by
This is because is independent of .
Let be a second order tensor. Then
Therefore,
Here is the fourth order identity tensor. In index notation with respect to an orthonormal basis
This result implies that
where
Therefore, if the tensor is symmetric, then the derivative is also symmetric and we get
where the symmetric fourth order identity tensor is
Let and be two second order tensors, then
In index notation with respect to an orthonormal basis
We also have
In index notation
If the tensor is symmetric then
Proof 

Recall that
Since , we can write Using the product rule for second order tensors we get or, Therefore, 
