Value of a determinant involving logarithmic, algebraic and trigonometric functions If $D(x) = \begin{vmatrix}
x&(1+x^2)&x^3\\
\log(1+x^2)&e^x&\sin(x)\\
\cos(x)&\tan(x)&\sin^2(x)\\
\end{vmatrix}$, then
find the correct option.
(a) $D(x)$ is divisible by $x$
(b) $D(x) = 0$
(c) $\frac{d}{dx}D(x) = 0$
(d) None of these
I just found that the correct option is (a). 
My attempt: Elaborate expansion of the given determinant is not helping towards the solution. I wonder if there is an elegant way to solve this question. Any hint would be very helpful.
 A: I would say (d)(a).


*

*To check "divisibility" by $x$, you'll ignore the terms which is the multiple of $x^m$ and focus on terms which don't seem to be a multiple of $x$, expand it to find out the divisibility.  In this case, expand the determinant along the first row, and only look at $1+x^2$.  The corresponding minor matrix doesn't not have any entries which are polynomial with respect to the variable $x$, so it's in the form $\require{cancel}\cancel{(\textit{elementary functions})x^2+(\textit{elementary functions})\cdot1}$, which is not a multiple of $x$.  Observe that the third column is "divisible" by $x$ in the sense of analytic functions  (as @Hans point out) since $\sin x=x+o(x^2)$.

*I'm going to disprove (c).  Once we have disproved (c), we will know that $D(x)$ is not a constant, so (b) will be wrong.

*Since you ask for "elegant" solution, explicit numerical solution won't satisfy you.  As a result, I invite you to think the behaviour of $|D(x)|$ as $x \to \pi/2^-$. ($x < \pi/2$ and $x\to\pi/2$) Except $\tan x$, the entries of the matrix are continuous at $\pi/2$, so is bounded as $x \in (\pi/2-\delta,\pi/2)$ for some small $\delta>0$.  Consider the expansion of the determinant along the third row.  The term containing $\tan(x)$ goes to $\infty$, while the other two terms are bounded.  From this, we conclude that $|D(x)|\to\infty$ as $x\to\pi/2^-$, so $D(x)$ is not constant.

*That's the only remaining choice. (a) is correct in the sense of analytic functions.

A: a) By substituting $\sin x=x\text{ sinc}(x)$, where $\text{sinc}$ is everywhere finite, you can pull out a factor $x$.
b, c) For $x=k\pi$, the matrix is triangular and the product of the diagonal is not constant*.
d) See a).

*Piecewise constant would rule out c) anyway.
