If $y = \begin{vmatrix} f(x) & g(x) & h(x) \\ l & m & n \\ a & b & c \end{vmatrix}$,prove that $\frac{dy}{dx} = \begin{vmatrix} f'(x) & g'(x) & h'(x) \\ l & m & n \\ a & b & c \end{vmatrix}$.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) Given $y = \begin{vmatrix} f(x) & g(x) & h(x) \\ l & m & n \\ a & b & c \end{vmatrix}$.
Expanding the determinant along the first row,we get:
$y = f(x)(mc - nb) - g(x)(lc - na) + h(x)(lb - ma)$.
Differentiating both sides with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}[f(x)(mc - nb)] - \frac{d}{dx}[g(x)(lc - na)] + \frac{d}{dx}[h(x)(lb - ma)]$.
Since $l, m, n, a, b, c$ are constants:
$\frac{dy}{dx} = f'(x)(mc - nb) - g'(x)(lc - na) + h'(x)(lb - ma)$.
This expression is the expansion of the determinant where the first row is differentiated:
$\frac{dy}{dx} = \begin{vmatrix} f'(x) & g'(x) & h'(x) \\ l & m & n \\ a & b & c \end{vmatrix}$.
Hence,the result is proved.

Explore More

Similar Questions

The determinant $\left| \begin{array}{ccc} 4 + x^2 & -6 & -2 \\ -6 & 9 + x^2 & 3 \\ -2 & 3 & 1 + x^2 \end{array} \right|$ for $x \neq 0$ is not divisible by:

If $S_{r} = \left|\begin{array}{ccc} 2r & x & n(n+1) \\ 6r^{2}-1 & y & n^{2}(2n+3) \\ 4r^{3}-2nr & z & n^{3}(n+1) \end{array}\right|$, then the value of $\sum_{r=1}^{n} S_{r}$ is independent of

Which of the following statements is false?
$1$. If $A$ is a skew-symmetric matrix of order $5 \times 5$,then the rank of $A$ is less than $5$.
$2$. If $P$ is a non-zero column matrix and $Q$ is a non-zero row matrix,then the rank of $PQ$ is $1$.
$3$. The rank of $\begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 4 \\ 5 & 6 & 7 \end{bmatrix}$ is $2$.
$4$. If the lines $a_r x + b_r y + c_r = 0$ $(r = 1, 2, 3)$ are distinct and intersect at a point,then the rank of $\begin{bmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{bmatrix}$ is $3$.

The rank of the matrix $\begin{bmatrix} 3 & 5 & -1 & 4 \\ 2 & 1 & 3 & -2 \\ 8 & 11 & 1 & 6 \\ -7 & -14 & 6 & -14 \end{bmatrix}$ is

In a matrix $A$,if all the sub-matrices of order $k$ are singular and there is at least one non-singular sub-matrix of order $r$ $(r < k)$,then the rank $(\rho)$ of the matrix $A$:

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo