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Differentiation of Determinants - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

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If $f(x)=\left|\begin{array}{ccc}2 \cos ^4 x & 2 \sin ^4 x & 3+\sin ^2 2 x \\ 3+2 \cos ^4 x & 2 \sin ^4 x & \sin ^2 2 x \\ 2 \cos ^4 x & 3+2 \sin ^4 x & \sin ^2 2 x\end{array}\right|$, then $\frac{1}{5} f^{\prime}(0)=$ is equal to:

If $f(x)=\left|\begin{array}{ccc}x^3 & 2 x^2+1 & 1+3 x \\ 3 x^2+2 & 2 x & x^3+6 \\ x^3-x & 4 & x^2-2\end{array}\right|$ for all $x \in \mathbb{R}$, then $2 f(0)+f^{\prime}(0)$ is equal to

Concepts Covered - 1

Differentiation of Determinants

Differentiation of Determinants

To differentiate a determinant, we differentiate one row or column at a time, keeping the other row or column unchanged.

Consider $2 \times 2$ matrix,
$\Delta=\left|\begin{array}{ll}y_{11} & y_{12} \\ y_{21} & y_{22}\end{array}\right|$, where $y_{i j}$ is a function of $x$ for all $i, j$
$\Delta=y_{11} \cdot y_{22}-y_{12} \cdot y_{21}$
differentiating w.r.t. $x$ we get

$
\begin{aligned}
\Delta & =\left(y_{11}\right)^{\prime} \cdot y_{22}+y_{11} \cdot\left(y_{22}\right)^{\prime}-\left(y_{12}\right)^{\prime} \cdot y_{21}-y_{12} \cdot\left(y_{21}\right)^{\prime} \\
& =\left[\left(y_{11}\right)^{\prime} \cdot y_{22}-\left(y_{12}\right)^{\prime} \cdot y_{21}\right]+\left[y_{11} \cdot\left(y_{22}\right)^{\prime}-y_{12} \cdot\left(y_{21}\right)^{\prime}\right] \\
& =\left|\begin{array}{cc}
\left(y_{11}\right)^{\prime} & \left(y_{12}\right)^{\prime} \\
y_{21} & y_{22}
\end{array}\right|+\left|\begin{array}{cc}
y_{11} & y_{12} \\
\left(y_{21}\right)^{\prime} & \left(y_{22}\right)^{\prime}
\end{array}\right|
\end{aligned}
$
Thus, for

$
\Delta=\left|\begin{array}{l}
R_1 \\
R_2
\end{array}\right|, \quad \Delta^{\prime}=\left|\begin{array}{c}
\left(R_1\right)^{\prime} \\
R_2
\end{array}\right|+\left|\begin{array}{c}
R_1 \\
\left(R_2\right)^{\prime}
\end{array}\right|
$

We can also differentiate column-wise.

Similarly, if a three-order determinant is given, 

$
\begin{aligned}
\Delta= & \left|\begin{array}{lll}
f(x) & g(x) & h(x) \\
p(x) & q(x) & r(x) \\
u(x) & v(x) & w(x)
\end{array}\right| \text {, then } \\
\Delta^{\prime}= & \frac{d}{d x}(\Delta)=\left|\begin{array}{lll}
f^{\prime}(x) & g^{\prime}(x) & h^{\prime}(x) \\
p(x) & q(x) & r(x) \\
u(x) & v(x) & w(x)
\end{array}\right|+\left|\begin{array}{ccc}
f(x) & g(x) & h(x) \\
p^{\prime}(x) & q^{\prime}(x) & r^{\prime}(x) \\
u(x) & v(x) & w(x)
\end{array}\right| \\
& +\left|\begin{array}{ccc}
f(x) & g(x) & h(x) \\
p(x) & q(x) & r(x) \\
u^{\prime}(x) & v^{\prime}(x) & w^{\prime}(x)
\end{array}\right|
\end{aligned}
$
Also if $\Delta=\left|\begin{array}{ccc}f(x) & g(x) & h(x) \\ a & b & c \\ p & q & r\end{array}\right|$, where, a,b,c,p,q and r are constant, then

$
\Delta^{\prime}=\left|\begin{array}{ccc}
f^{\prime}(x) & g^{\prime}(x) & h^{\prime}(x) \\
a & b & c \\
p & q & r
\end{array}\right|
$

and $\quad \frac{d^n}{d x^n}(\Delta)=\left|\begin{array}{ccc}\frac{d^n}{d x}(f(x)) & \frac{d^n}{d x}(g(x)) & \frac{d^n}{d x}(h(x)) \\ a & b & c \\ p & q & r\end{array}\right|$

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Differentiation of Determinants

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Differentiation of Determinants

Mathematics for Joint Entrance Examination JEE (Advanced) : Calculus

Page No. : 3.17

Line : 18

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