Prove $l_{ip}=\frac{\partial \bar x_i}{\partial x_p}=\frac{\partial x_p}{\partial \bar x_i}$ Let $l$ be a rotation tensor such that 
$$\bar x_i=l_{ip}x_p$$ 
where $l_{ip}$ is the direction cosine between the unit vectors in the component directions $x_p$ and $\bar x_i$. Prove that 
$$l_{ip}=\frac{\partial \bar x_i}{\partial x_p}=\frac{\partial x_p}{\partial \bar x_i}$$ 

Hint 
Use this property of the rotation tensor
$$l{l^T} = I$$
 A: OK, let us start from
$${\bar x_i} = {l_{ip}}{x_p}\tag{1}$$
Now, we take the derivative with respect to $x_k$ to obtain
$${{\partial {{\bar x}_i}} \over {\partial {x_k}}} = {l_{ip}}{{\partial {x_p}} \over {\partial {x_k}}} = {l_{ip}}{\delta _{pk}} = {l_{ik}}\,\,\,\, \to \,\,\,\,{l_{ip}} = {{\partial {{\bar x}_i}} \over {\partial {x_p}}}\tag{2}$$
Next, we note that $l_{ip}$ has the following property that its inverse is equal to its transpose. This is called orthogonality. Hence, we can write this property in index form as
$$\begin{array}{*{20}{l}}
{l{l^T} = I}& \to &{l = {{\left( {{l^{ - 1}}} \right)}^T}}\\
{{l_{ip}}{l_{kp}} = {\delta _{ik}}}& \to &{{l_{ip}} = {{\left( {{l^{ - 1}}} \right)}^T}_{ip}}
\end{array}\tag{3}$$
On the other hand, we have
$$\eqalign{
  & {\left( {{l^{ - 1}}} \right)_{ip}} = {{\partial {x_i}} \over {\partial {{\bar x}_p}}}  \cr 
  & {\left( {{l^{ - 1}}} \right)^T}_{ip} = {{\partial {x_p}} \over {\partial {{\bar x}_i}}} \cr}\tag{4}$$
Finally, combine $(2)$, $(3)$, and $(4)$ to get
$${l_{ip}} = {{\partial {{\bar x}_i}} \over {\partial {x_p}}} = {{\partial {x_p}} \over {\partial {{\bar x}_i}}}\tag{5}$$
