Application of Rolle's theorem and determinant $f(x) =
\begin{vmatrix}
 \sin^3x & \sin^3 a & \sin^3 b \\ 
 xe^x & ae^a & be^b\\
\frac{x}{1+x^2} & \frac{a}{1+a^2} & \frac{b}{1+b^2}
\end{vmatrix}$
Where $0<a<b<2π$ then show that the equation $f^{'}(x)=0$  has atlest one root in the interval $(a,b)$
My approach
First try to differentiate the give determinant and find that $f^{'}(x)=0$ then I apply the Rolle's theorem to solve the question. 
I face the difficulty in differentiate the determinant. 
 A: I am quoting a rule to differentiate a determinant:
"You can differentiate the first row (or column) and keep the entries of the other row untouched. Then you can get a determinant. Do this for all other rows (or columns). Then find the sum of all such determinants."
Applying this rule here,  
f'(x) =
\begin{vmatrix}
3 \sin^2x \cos x & \sin^3 a & \sin^3 b \\ 
e^x + xe^x & ae^a & be^b\\
\frac{1-x^2}{(1+x^2)^2} & \frac{a}{1+a^2} & \frac{b}{1+b^2}
\end{vmatrix}
However, this does not solve your problem. Let's look at the statement of Rolle's theorem:
If a real-valued function $f$ is continuous on a proper closed interval $[a, b]$, differentiable on the open interval $(a, b)$, and $f (a) = f (b)$, then there exists at least one $c$ in the open interval $(a, b)$ such that
${\displaystyle f'(c)=0}$
Notice that in this case, $f(a) = f(b)$, right? [If you put $x = a$ or $x = b$, two columns become identical and the determinant vanishes]. That is all you need!
A: Let $\mathbf{c}$ be the vector-valued function which takes $x$ to $\left<\sin^3x,xe^x,x/(1+x^2)\right>$.  Then $f(x)$ is the scalar triple product of $\mathbf{c}(x)$, $\mathbf{c}(a)$, and $\mathbf{c}(b)$:
$$
    f(x) = \mathbf{c}(x) \cdot \left(\mathbf{c}(a) \times \mathbf{c}(b)\right)
$$
Because the cross product of two vectors in $\mathbb{R}^3$ is orthogonal to each of the individual vectors, $f(a) = f(b) = 0$.  So by Rolle's Theorem there exists a point $x$ between $a$ and $b$ where $f'(x) = 0$.
