Question

If $f\left(x\right)=\left|\begin{array}{ccc}x& x^2& x^3\\ 1& 2x& 3x^2\\ 0& 2& 6x\end{array}\right|$, then find $f^{\prime }\left(x\right)$.

• A. $f^{\prime }\left(x\right)=5x^4$
• B. $f^{\prime }\left(x\right)=6x^3$
• C. $f^{\prime }\left(x\right)=5x^2$
• D. $f^{\prime }\left(x\right)=6x^2$

Answer 1

The correct option is D $f^{\prime }\left(x\right)=6x^2$

$f\left(x\right)=\left|\begin{array}{ccc}x& x^2& x^3\\ 1& 2x& 3x^2\\ 0& 2& 6x\end{array}\right|$
On differentiating, we get

$f\left(x\right)=\left|\begin{array}{ccc}\frac{d}{dx}\left(x\right)& \frac{d}{dx}\left(x^2\right)& \frac{d}{dx}\left(x^3\right)\\ 1& 2x& 3x^2\\ 0& 2& 6x\end{array}\right|+\left|\begin{array}{ccc}x& x^2& x^3\\ \frac{d}{dx}\left(1\right)& \frac{d}{dx}\left(2x\right)& \frac{d}{dx}\left(3x^2\right)\\ 0& 2& 6x\end{array}\right|+\left|\begin{array}{ccc}x& x^2& x^3\\ 1& 2x& 3x^2\\ \frac{d}{dx}\left(0\right)& \frac{d}{dx}\left(2\right)& \frac{d}{dx}\left(6x\right)\end{array}\right|$

or $f^{\prime }\left(x\right)=\left|\begin{array}{ccc}1& 2x& 3x^2\\ 1& 2x& 3x^2\\ 0& 2& 6x\end{array}\right|+\left|\begin{array}{ccc}x& x^2& x^3\\ 0& 2& 6\\ 0& 2& 6x\end{array}\right|+\left|\begin{array}{ccc}x& x^2& x^3\\ 1& 2x& 3x^2\\ 0& 0& 6x\end{array}\right|$

As we know, if any two rows or columns are equal, then value of the determinant is zero.

$=0+0+\left|\begin{array}{ccc}x& x^2& x^3\\ 1& 2x& 3x^2\\ 0& 0& 6x\end{array}\right|$
$\therefore f^{\prime }\left(x\right)=6(2x^2-x^2)$

Therefore, $f^{\prime }\left(x\right)=6x^2$