1. Tardigrade
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  3. Mathematics
  4. If y = |f(x)&g(x)&h(x) 1&m&n a&b&c|, then (dy/dx) is equal to

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Q. If $y = \begin{vmatrix}f\left(x\right)&g\left(x\right)&h\left(x\right)\\ 1&m&n\\ a&b&c\end{vmatrix}$, then $\frac{dy}{dx}$ is equal to

KCETKCET 2017Determinants

Solution:

We have,
$y=\begin{bmatrix}f(x) & g(x) & h(x) \\ 1 & m & n \\ a & b & c\end{bmatrix}$
$\therefore \frac{dy}{dx}=\begin{bmatrix}\frac{d}{d x} f(x) & \frac{d}{d x} g(x) & \frac{d}{d x} h(x) \\ l & m & n \\ a & b & c\end{bmatrix}$
$+\begin{bmatrix}f(x)& g(x) & h(x) \\ \frac{d}{d x} & \frac{d}{d x} m & \frac{d}{d x} \\ a & b & c\end{bmatrix}+\begin{bmatrix}f(x) & g(x) & h(x) \\ 1 & m & n \\ \frac{d}{d x} a & \frac{d}{d x} b & \frac{d}{d x} c\end{bmatrix}$
$=\begin{bmatrix}f'(x) & g'(x) & h'(x) \\l & m & n \\ a & b & c\end{bmatrix}+\begin{bmatrix}f(x) & g(x) & h(x) \\ 0 & 0 & 0 \\ a & b & c\end{bmatrix}$
$+\begin{bmatrix}f(x) & g(x) & h(x) \\ l & m & n \\ 0 & 0 & 0\end{bmatrix}=\begin{bmatrix}f'(x) & g'(x) & h'(x) \\ l & m & n \\ a & b & c\end{bmatrix}$
$=\begin{bmatrix} & m & n \\ a & b & c \\ f'(x) & g'(x) & h'(x)\end{bmatrix}=\begin{bmatrix}f'(x) & 1 & a \\ g'(x) & m & b \\ h'(x) & n & c\end{bmatrix}$