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

Concept: Determinant Evaluation Using Identities — differentiating a determinant by differentiating only one row at a time.

Step 1: Write $y$ as a sum of three determinants, each with one row differentiated, while the other rows stay unchanged. This follows from the product rule applied to the expansion of a determinant.

Step 2: Since the second and third rows contain constants ($l, m, n$ and $a, b, c$), their derivatives are zero. Therefore, the determinants where those rows are differentiated vanish. …

We start with the given determinant:

Here, $l, m, n, a, b, c$ are constants — they do not depend on $x$. Only the first row contains functions of $x$. The question asks us to prove that differentiating $y$ with respect to $x$ gives the same determinant, but with the first row replaced by $f'(x), g'(x), h'(x)$.

The core idea is simple: a determinant is a sum of products of entries, each product containing exactly one entry from each row and each column. When you differentiate such a sum term-by-term, the product rule applies. Since only the first row's entries vary with $x$, any term's derivative will involve the derivative of exactly one factor from that row — the other two rows are constant and come along for the ride. The result reorganises itself neatly into the determinant of the derivative row.

Let's walk through it step by step.

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1. Write the determinant in expanded form. Using the standard expansion along the first row:

Each $2 \times 2$ determinant is a constant (since all its entries are constants). Let’s denote them:

So $y = C_1 f(x) - C_2 g(x) + C_3 h(x)$.

2. Differentiate term by term. Since $C_1, C_2, C_3$ are constants:

3. Recognise this as a determinant again.

   The expression $C_1 f'(x) - C_2 g'(x) + C_3 h'(x)$ is exactly the expansion of:

   $\begin{vmatrix} f'(x) & g'(x) & h'(x) \\ l & m & n \\ a & b & c \end{vmatrix}$ …

Method: Differentiating a Determinant Row-by-Row

When only one row (or column) of a determinant depends on $x$ and the rest are constants, the derivative of the determinant is found by differentiating only that row, leaving the others unchanged — this follows from expanding the determinant along that row and applying the product rule to each term.

Steps

Step 1: Expand the determinant along the row that depends on $x$

Write it as a sum of each entry in that row times its (constant) cofactor.

Step 2: Recognise that each cofactor is a constant (built only from the other, constant rows)

Step 3: Differentiate the expanded sum — since the cofactors are constant, only the entries themselves get differentiated

Step 4: Recognise the differentiated sum as the expansion of a new determinant with the original row replaced by its derivative …

Common Mistakes

Mistake 1: Attempting to differentiate every row of the determinant, not just the row that depends on $x$.

Why it's wrong: the second and third rows here ($l,m,n$ and $a,b,c$) are all constants — differentiating them would give zero rows and produce a wrong (degenerate) result; if more than one row genuinely depended on $x$, the derivative would instead be a SUM of several determinants, each with a different row replaced. Correct approach: identify which row(s) actually contain functions of $x$ before differentiating anything.

Mistake 2: Trying to expand the full $3\times3$ determinant symbolically and differentiate the entire expression at once, rather than recognising the cofactors as constants. …