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Some JEE questions hand you a 3×3 determinant full of functions of x and ask for a limit, a derivative or a maximum. Expanding first works, but it is slow and error-prone. There is a quicker rule, and once you see why it works you will spot the questions built around it.
How do you differentiate a determinant?
The rule, as this Queen's College, Hong Kong teaching note puts it: differentiate the first row and keep the other rows untouched, do the same for every other row, then add all the determinants you get.
For a 3×3 determinant f(x) with rows R1, R2 and R3:
f′(x) = |R1′, R2, R3| + |R1, R2′, R3| + |R1, R2, R3′|
the same rule works column by column, if the functions sit in a column
it works for any size: an n×n determinant gives n terms
Why does this work? A determinant is linear in each row while the other rows stay fixed, so the product rule sends the derivative into one row at a time. The fully general version is Jacobi's formula, d/dt det A(t) = tr(adj(A(t)) dA/dt), named after the mathematician Carl Jacobi. One of the proofs in the Wikipedia article on Jacobi's formula starts from Laplace's cofactor expansion. Written out using cofactors, it gives the same row-by-row sum.
What happens when a row is constant?
If a row holds only numbers, its derivative is a row of zeros, and a determinant with a zero row equals zero. That term disappears. Many JEE-style questions are built this way: two rows of constants and one row of functions means f′(x) is a single determinant.
A second shortcut: if differentiating a row gives you a copy of another row, that term is zero too, because two equal rows make a determinant zero.
Try it on f(x) = |x, x², x³ ; 1, 2x, 3x² ; 0, 2, 6x|, where semicolons separate the rows.
Differentiate R1: you get (1, 2x, 3x²), which is R2. Equal rows, so this term is 0.
Differentiate R2: you get (0, 2, 6x), which is R3. Zero again.
Differentiate R3: you get (0, 0, 6). This is the only term left: |x, x², x³ ; 1, 2x, 3x² ; 0, 0, 6| = 6(2x² − x²) = 6x².
Check by expanding: f(x) = 2x³, so f′(x) = 6x². It matches. Differentiating every entry at once would have given 12, which is wrong.
How does JEE use this for limits?
When f(0) = 0, the limit of f(x)/x as x → 0 is just f′(0), the definition of the derivative. The rule means you never have to expand.
Take f(x) = |sin x, 1 − cos x, x ; 1, 0, 2 ; 3, 1, 4|. Row 1 is zero at x = 0, so f(0) = 0. Rows 2 and 3 are constant, so only row 1 is differentiated:
f′(x) = |cos x, sin x, 1 ; 1, 0, 2 ; 3, 1, 4|
f′(0) = |1, 0, 1 ; 1, 0, 2 ; 3, 1, 4| = 1(0 − 2) − 0 + 1(1 − 0) = −1
So the limit is −1. For why a limit like this is a derivative in disguise, our guide to limits and continuity for JEE covers the standard forms.
How do you find a maximum when the function is a determinant?
The method is the usual one: set f′(x) = 0, then check the sign of f″(x). The determinant rule just gives you f′ and f″ faster.
Let f(x) = |x², x, 1 ; 1, 2, 3 ; 0, 1, 1|.
Only R1 varies, so f′(x) = |2x, 1, 0 ; 1, 2, 3 ; 0, 1, 1| = 2x(2 − 3) − 1(1 − 0) = −2x − 1.
f′(x) = 0 gives x = −1/2.
Differentiate again: f″(x) = |2, 0, 0 ; 1, 2, 3 ; 0, 1, 1| = 2(2 − 3) = −2, which is negative, so x = −1/2 is a maximum.
Expanding the original gives f(x) = −x² − x + 1, so the maximum value is f(−1/2) = 5/4.
For the wider set of first- and second-derivative tests, see maxima, minima and rates of change.
A routine to practise this week
Rework the three examples above on paper without looking.
Take five determinant questions from past JEE papers and, before solving, write down which terms will vanish.
Time yourself with the rule, then by expanding, and compare.
Revise the properties that make terms vanish (zero rows, equal rows, row operations) in our post on matrices and determinants for JEE.
If a step goes wrong and you cannot see where, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer. If the gap is in the idea itself, multilinearity or cofactors, the Fermi Concept Library gives each mathematics concept its own lesson.
