For the square root, already, one sees that, if the matrix isn't positive definite, there's a problem. So absent any further information, not much can be said. If one is looking for a matrix such that A^n = B, given B, then things can get very complicated very fast. A lot depends on the specific problem, because the matrices one is interested in manipulating this way, usually, have some additional properties.

First of all it is nonsense to ask  what are the roots of a matrix?  In order to establish an algebra in given set of objects you have to define summation and multiplication operations.  We define a matrix  by specifying the number of rows M and columns  N, and the we called  as MXN  matrix. The set of  M X N  matriices forms a class, and you can define simple addition operation among the corresponding  elements of each members,   which it turns,  obeys  commutative and associative rules. This set  is a group,  and it is not an algebra since you can not define  multiplication operation unless its members are comformable in the  order   stated, namely;

 M x N multiply  N x Z =   M x Z,     The class of conformable matrices forms  also  a group if you define pre, and post multiplication operatins  which is associative but not commutative.  As one may see that one can only form  a  noncommutative  algebra   out of set of square matrices composed of matrices having same dimensions (order).  Now, you may  established calculus  if you define a continous set of  conformable square matrices having infinite members,  usual way by applying the  calculus to each elements of a given matrix in the set. NO PROBLEM!!.

There is close connection between the solution of a  linear set of  algebraic equation, and the matrix equation denoted as   A X = B.  There is an unique solution if only if the determinant of the coefficient matrix is nonzero (not singular). İf it is not singular you can still apply to see the whole scenario {'ranks of coefficient and augmented matrices, etc, ! }Then you can apply  CRAMER's rule systematic fashion to get the n-roots, which may not be distinct!  Actually at the same time you can get the inverse matrix as well. There are many techniques to reduce the computation steps or time  in practice. For large matrices  [1000 x 1000]   this systematic doesn' t operate we have other methods to deal with.

Biggest problem comes in when you deal with the asymmetric matrices to find their  eigenvalues and eigenvectors. Even though we know that there are equal number of eigenvalues as their orders.  There is no way of saying how many distinct eigenvectors  are there!  if there is  a degeneracy!!! 

It depends on the type of derivative, if it is the calssical derivative just take the derivative of entrie. For the derivative .

For the derivitaive of the nth root of A , start by computing log(A) by using Richter formula or other theniques (see http://epubs.siam.org/doi/abs/10.1137/1.9780898717778.ch11) then use the exponential

It is singular but the ranks of  the coefficient and augmented matrices are equal then the solutions are  exist having  (N- r)  degrees of freedom represented by   N-r  linearly independent eigenvectors, which they come out as a by product. Otherwise no solution exist, and this  set of linear inhomogeneous equations (B NEQ  0) are called inconsistent or incompatible,

I also suggest you read some papers of Prof. Nicholas J. Higham. He is an expert on this topic, you can refer to his homepage: 

http://www.ma.man.ac.uk/~higham/

All these are text book knowledge. I have been teaching that topic the  last 40 years as a single chapter from Hildebrand text book of MIT,  which is included  as a reference for a chapter dealing with linear algebra only in the curriculum of two semesters course  on applied mathematics for graduate students at METU.

Francis Hildebrand @ Methods of Applied Mathematics. 1952.  

You can find information on your question in my papers "EVALUATING THE FRECHET DERIVATIVE OF THE MATRIX PTH ROOT" and "COMPUTATION OF THE MATRIX PTH ROOT AND ITS FRECHET DERIVATIVE BY INTEGRALS", that can be easily downloaded from my researchgate page. 

Dear Morteza Noruzi,

1)  If  A(t) = [a_{ij}(t)], then  A'(t) = [d/dt(a_{ij}(t))].

2)  A real positive semidefinite symmetric matrix  A  has a real square root. By using Spectral Decomposition.

     An complex positive semidefinite (Hermitian) matrix A in M_{n} (written as A \succeq 0), there exists a unique B \succeq 0 so that B^{2} = A, that is  A  has a unique square root.

See for example:

W.-H. Steeb, Problems and solutions in introductory and advanced matrix calculus, World Scientific Publishing Co. Pte. Ltd., Toh Tuck Link, Singapore, 2006. pp.89-90

Using Newton's Method applied to X^p=A you can obtain good solutions, also can use another iteratives methods developed  to ensure stability.

See the book of Nick Higham, Functions of a matrix. Theory and Computation , 2008.

The derivative of a Matrix is the matrix of the derivatives . It is enough to calculate the derivative of each term so to create the derivative Matrix.

For calculating the nth root of a Matrix, you may use the same procedure that you use to make the exponential or logarithm just applying the function root instead of using exponential or logarithm. The procedure is sometimes called operational calculus and you can find it in the Book

Piccinini, L. C./Stampacchia, G./Vidossich, G., Ordinary Differential Equations in Rn. Problems and Methods. Berlin-Heidelberg-New York-Tokyo, Springer-Verlag 1984. XII, 385 S., 38 Abb., DM 96,  ISBN 3-540-90723-8.

For square root in small dimension, you can also solve directly the equation systems (unfortunately they are not linear).

For further information please write on the RG to Livio Clemente Piccinini.

It is not so difficult. All you need is to work with the computation of a polynomial for a matrix. Take a first look here:

http://www.millersville.edu/~bikenaga/linear-algebra/matrix-exponential/matrix-exponential.html

Then try to read this:

http://www.cs.cornell.edu/cv/researchpdf/19ways+.pdf