Spivak's Calculus on Manifolds problem 2-15(c) 
2-15. Regard an $n \times n $ matrix as a point in the $n$-fold product
   $\Bbb{R^n} \times \cdots\times \Bbb{R^n}$ by considering each row as a
   member of $\Bbb{R^n}$.
(a) Prove that $det: \Bbb{R^n} \times \cdots \times \Bbb{R^n} \to
\Bbb{R}$ is differentiable and $
 D(\mathrm{det})(a_1,\ldots,a_n)(x_1,\ldots,x_n)=\sum_{n=i}^{n} \mathrm{det} \begin{bmatrix} 
a_1 \\ \vdots \\ x_i\\ \vdots\\ a_n \end{bmatrix}$
(b) if $a_{ij}: \Bbb{R} \to \Bbb{R} $ are differentiable and
   $f(t)=det(a_{ij}(t))$, show that $f'(t)= \sum_{j=1}^{n} det
\begin{bmatrix}  a_{11}(t),\ldots, a_{1n}(t)\\ \vdots\\ a_{j1}'(t),\ldots,
a_{jn}'(t)\\ \vdots \\ a_{n1}(t),\ldots, a_{nn}(t) \end{bmatrix}$
(c) if $\mathrm{det} (a_{ij}(t)) \neq 0$ for all $t$ and $b_1,...,b_n: \Bbb{R}
\to \Bbb{R}$ are differentiable, let $s_1,\ldots,s_n: \Bbb{R}
 \to \Bbb{R}$ be the functions such that $s_1(t),\ldots,s_n(t)$
   are the solutions of the equations $\sum_{j=1}^{n}
 a_{ij}(t)s_j(t)=b_i(t)$ $i=1,\ldots,n$.
show that $s_i$ is differentiable and find $s_i'(t)$.

Problem 2-40 asks to redo problem 2-15(c) using the implicit function theorem.

2-12 theorem (implicit function theorem). Suppose $ f: \Bbb{R^n}
\times \Bbb{R^m} \to \Bbb{R^m} $ is continuously
  differentiable in an open set containing $(a,b)$ and $f(a,b) =0$. Let  be the $m \times m$ matrix. 
  $$(D_{n+j}f^i(a,b))~, \quad 1 \leq i,~j \leq m~.$$
  If $\mathrm{det} M \neq 0$, there is an open set $A \subset R^n$ containing $a$
  and an open set $B$ subset $R^m$ containing $b$, with the following
  property: for each $x \in A$ there is a unique $g(x)$ in $B$ such that $f(x,g(x)) =0$. the function $g$ is differentiable.

Call $i^{th}$ row of $(a_{ji}(t))$ as $R_i(t)$. Let me define $g(t)= 
\begin{bmatrix} 
s_1(t) \\
s_2 (t)\\
\vdots   \\
s_n(t) 
\end{bmatrix}$. thus $R_i(t)g(t)=b_i(t)$. If I define $f:\Bbb{R^n} \times \Bbb{R^n} \to \Bbb{R^n}$ such that $f(R_i(t),g(t)) = R_i(t)g(t)=b_i(t)$, the first difficulty I face is that the theorem will provide $g(t)$ is differentiable only if $b_i(t)=0$, which may not be true.
To overcome this difficulty, how should I define $f$ so that the theorem will provide me that $g(t)$ is differentiable?
 A: We have the system of linear equations 
$\begin{pmatrix}
a_{11}(t) &  a_{12}(t) & \cdots & a_{1n}(t) \\
a_{21}(t) &  a_{22}(t) & \cdots & a_{2n}(t) \\
\vdots &  \vdots & \ddots & \vdots \\
a_{n1}(t) &  a_{n2}(t) & \cdots & a_{nn}(t) \\
\end{pmatrix} \begin{pmatrix}
s_1(t) \\ s_2(t) \\ \vdots \\ s_n(t)
\end{pmatrix}= \begin{pmatrix}
b_1(t) \\ b_2(t) \\ \vdots \\ b_n(t)
\end{pmatrix}$
Which is convenient to write in matrix form
$A(t)s(t) = b(t)$.
The goal is to show that if the matrix $A(t)$ is invertible for all $t$, then solutions exist, are differentiable (and the book describes how one may go about computing the derivatives).
To set-up the application of the Implicit function theorem, we can define
$f : \mathbb{R} \times \mathbb{R}^n \rightarrow \mathbb{R}^n$
by 
$f(t,\, s) = A(t)s - b(t)$.
This completes the set-up of the question, which is all you have asked, but I will continue for my own curiosity.

Now we should look at the matrix
$(D_{1+j}f^i(t,\,s))_{1 \leq i,j \leq n}$
We have 
$f^i(t,\,s) = \left(\sum_{k=1}^n a_{ik}(t)s_k\right) - b_i(t)$
and 
$\begin{align*}
D_{1+1}f^i(t,\,s) &= \frac{\partial f^i(t,\,s)}{\partial s_1} = a_{i1}(t) \\
D_{1+2}f^i(t,\,s) &= \frac{\partial f^i(t,\,s)}{\partial s_2} = a_{i2}(t) \\
\vdots \\
D_{1+n}f^i(t,\,s) &= \frac{\partial f^i(t,\,s)}{\partial s_n} = a_{in}(t) 
\end{align*}$
So the matrix in the statement of the I.F.T. is just
$(D_{1+j}f^i(t,\,s))_{1 \leq i,j \leq n} = A(t)$.
By hypothesis on $A(t)$, this is invertible for all $t$. In particular, about any $t_0$, there is an open set $I$ containing $t_0$ and an open set $B \subseteq \mathbb{R}^n$ containing $s_0$, such that for each $t \in I$, there is a unique $s(t) \in B$  such that $0 = f(t,\,s(t)) = A(t)s(t) - b(t)$, i.e. $s(t)$ is a solution of the system of linear equations, and $s(t)$ is a differentiable function of $t$.
