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Chapter 5 · Continuity and Differentiability

Miscellaneous · Q21: Derivative of a determinant whose first row is f,g,hf, g, h

Miscellaneous ExerciseDerivatives · Determinants · Miscellaneous Exercise
The question

If

y=∣f(x)g(x)h(x)lmnabc∣,y = \begin{vmatrix} f(x) & g(x) & h(x) \\ l & m & n \\ a & b & c \end{vmatrix},

prove that

dydx=∣f′(x)g′(x)h′(x)lmnabc∣.\frac{dy}{dx} = \begin{vmatrix} f'(x) & g'(x) & h'(x) \\ l & m & n \\ a & b & c \end{vmatrix}.
Exam answer

Expanding along R1R_1⁠,

y=f(x)(mc−nb)−g(x)(lc−na)+h(x)(lb−ma)\begin{aligned} y &= f(x)(mc - nb) \\ &\quad - g(x)(lc - na) \\ &\quad + h(x)(lb - ma) \end{aligned}

Since l,m,n,a,b,cl, m, n, a, b, c are constants, differentiating w.r.t. xx⁠,

dydx=f′(x)(mc−nb)−g′(x)(lc−na)+h′(x)(lb−ma)\begin{aligned} \frac{dy}{dx} &= f'(x)(mc - nb) \\ &\quad - g'(x)(lc - na) \\ &\quad + h'(x)(lb - ma) \end{aligned}

This is the expansion along R1R_1 of the determinant with first row f′(x),g′(x),h′(x)f'(x), g'(x), h'(x)⁠.

∴ dydx=∣f′(x)g′(x)h′(x)lmnabc∣\therefore\ \dfrac{dy}{dx} = \begin{vmatrix} f'(x) & g'(x) & h'(x) \\ l & m & n \\ a & b & c \end{vmatrix}⁠. Hence proved.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

Expand the determinant along the first row. Each of f(x)f(x)⁠, g(x)g(x)⁠, h(x)h(x) gets multiplied by a 2×22\times2 determinant built only from the constants l,m,n,a,b,cl, m, n, a, b, c⁠. So yy is just "constant times ff" plus "constant times gg" plus "constant times hh", and differentiating simply puts a prime on ff⁠, gg and hh⁠. Reading the result backwards, it is the expansion of the second determinant.

Given, to show, and the plan
Given
ff⁠, gg⁠, hh are differentiable functions of xx⁠; l,m,n,a,b,cl, m, n, a, b, c are constants (here a,b,ca, b, c are just names for the entries of the third row).
To show
dydx\dfrac{dy}{dx} equals the same determinant with the first row replaced by f′(x),g′(x),h′(x)f'(x), g'(x), h'(x)⁠.
Plan
(1) Expand yy along the first row. (2) Note the cofactors are constants. (3) Differentiate term by term. (4) Recognise the result as a determinant.

Basics you need

1Expanding a 3×33\times3 determinant along the first row

Each first-row entry times the 2×22\times2 determinant left after deleting its row and column, with signs +,−,++, -, + (Chapter 4, Section 4.2.3):

∣pqrlmnabc∣=p∣mnbc∣−q∣lnac∣+r∣lmab∣.\begin{aligned} \begin{vmatrix} p & q & r \\ l & m & n \\ a & b & c \end{vmatrix} &= p\begin{vmatrix} m & n \\ b & c \end{vmatrix} - q\begin{vmatrix} l & n \\ a & c \end{vmatrix} \\ &\quad + r\begin{vmatrix} l & m \\ a & b \end{vmatrix}. \end{aligned}
22×22\times2 determinants

Product of the main diagonal minus product of the other diagonal (Chapter 4, Section 4.2.2):

∣mnbc∣=mc−nb.\begin{vmatrix} m & n \\ b & c \end{vmatrix} = mc - nb.
3Sum and constant multiple rules

(u±v)′=u′±v′(u \pm v)' = u' \pm v' (Section 5.3, rule (1)), and (ku)′=ku′(ku)' = ku' for a constant kk (rule (2) with k′=0k' = 0⁠). Basics: derivative rules

Solution

  1. Expand yy along the first row

    y=f(x)∣mnbc∣−g(x)∣lnac∣+h(x)∣lmab∣.\begin{aligned} y &= f(x)\begin{vmatrix} m & n \\ b & c \end{vmatrix} - g(x)\begin{vmatrix} l & n \\ a & c \end{vmatrix} \\ &\quad + h(x)\begin{vmatrix} l & m \\ a & b \end{vmatrix}. \end{aligned}

    Evaluating the 2×22\times2 determinants:

    y=f(x)(mc−nb)−g(x)(lc−na)+h(x)(lb−ma).\begin{aligned} y &= f(x)(mc - nb) - g(x)(lc - na) \\ &\quad + h(x)(lb - ma). \end{aligned}
    Expanding along the first row puts the three functions outside the small determinants. Each 2×22\times2 determinant is what is left after deleting the row and column of that entry, and the signs follow the pattern +,−,++, -, +⁠.
    Determinant of order 3, expansion along R1R_1 (Chapter 4, Section 4.2.3; Basics 1, 2).

  2. The three multipliers are constants

    Write A1=mc−nbA_1 = mc - nb⁠, A2=−(lc−na)A_2 = -(lc - na)⁠, A3=lb−maA_3 = lb - ma⁠. Then

    y=A1f(x)+A2g(x)+A3h(x),y = A_1 f(x) + A_2 g(x) + A_3 h(x),

    where A1,A2,A3A_1, A_2, A_3 do not depend on xx⁠.

    They are built only from l,m,n,a,b,cl, m, n, a, b, c⁠, which are constants. (They are the cofactors of the first-row entries.) This is the key point: yy is a sum of constant multiples of ff⁠, gg and hh⁠.
    Cofactors (Chapter 4, Section 4.4).

  3. Differentiate term by term

    Since ff⁠, gg⁠, hh are differentiable and A1,A2,A3A_1, A_2, A_3 are constants,

    dydx=A1f′(x)+A2g′(x)+A3h′(x),\frac{dy}{dx} = A_1 f'(x) + A_2 g'(x) + A_3 h'(x),

    that is,

    dydx=f′(x)(mc−nb)−g′(x)(lc−na)+h′(x)(lb−ma).\begin{aligned} \frac{dy}{dx} &= f'(x)(mc - nb) - g'(x)(lc - na) \\ &\quad + h'(x)(lb - ma). \end{aligned}
    The sum rule lets us differentiate each term separately; the constant multiple rule keeps each AiA_i in front. No product rule is needed, because the other factor in each product is constant.
    Sum and constant multiple rules, Section 5.3 (Basic 3). Basics: derivative rules

  4. Recognise the result as a determinant

    Step 3's expression is exactly the first-row expansion (Basic 1) of the determinant whose first row is f′(x),g′(x),h′(x)f'(x), g'(x), h'(x) and whose other rows are unchanged:

    dydx=∣f′(x)g′(x)h′(x)lmnabc∣.\frac{dy}{dx} = \begin{vmatrix} f'(x) & g'(x) & h'(x) \\ l & m & n \\ a & b & c \end{vmatrix}.
    Expanding this new determinant along its first row gives f′(x)(mc−nb)f'(x)(mc - nb) −g′(x)(lc−na)- g'(x)(lc - na) +h′(x)(lb−ma)+ h'(x)(lb - ma)⁠, because its second and third rows, and so its cofactors A1,A2,A3A_1, A_2, A_3⁠, are the same as before. The two expressions are identical, which completes the proof.
    Expansion along R1R_1 read in reverse (Chapter 4, Section 4.2.3).
Result

For differentiable ff⁠, gg⁠, hh and constants l,m,n,a,b,cl, m, n, a, b, c⁠:

ddx∣f(x)g(x)h(x)lmnabc∣=∣f′(x)g′(x)h′(x)lmnabc∣,\begin{aligned} & \frac{d}{dx}\begin{vmatrix} f(x) & g(x) & h(x) \\ l & m & n \\ a & b & c \end{vmatrix} \\ &= \begin{vmatrix} f'(x) & g'(x) & h'(x) \\ l & m & n \\ a & b & c \end{vmatrix}, \end{aligned}

since both sides equal f′(x)(mc−nb)f'(x)(mc - nb) −g′(x)(lc−na)- g'(x)(lc - na) +h′(x)(lb−ma)+ h'(x)(lb - ma)⁠.

Check it with numbers

Take f=sin⁡xf = \sin x⁠, g=x2g = x^2⁠, h=exh = e^x⁠, second row 2,−1,32, -1, 3⁠, third row 1,4,51, 4, 5⁠. The cofactors are mc−nb=−5−12=−17mc - nb = -5 - 12 = -17⁠, lc−na=10−3=7lc - na = 10 - 3 = 7⁠, lb−ma=8+1=9lb - ma = 8 + 1 = 9⁠, so

y=−17sin⁡x−7x2+9ex.y = -17\sin x - 7x^2 + 9e^x.

At x=0.5x = 0.5⁠: y≈4.9382573y \approx 4.9382573⁠, and dydx=−17cos⁡x−14x+9ex≈−7.0804121\frac{dy}{dx} = -17\cos x - 14x + 9e^x \approx -7.0804121⁠. The determinant with first row cos⁡0.5,1,e0.5\cos 0.5, 1, e^{0.5} also evaluates to ≈−7.0804121\approx -7.0804121⁠, and a numerical derivative of yy (h=0.0001h = 0.0001⁠) agrees. ✓

Common mistakes
  1. Differentiating every entry. The constant rows would become rows of zeros and the determinant would be 0. Only the row containing functions is differentiated.
  2. Sign errors in the expansion: the middle term carries a minus sign, −g(x)(lc−na)-g(x)(lc - na)⁠.
  3. Calling a,b,ca, b, c variables. In this question they are constant entries.
What to take away
  1. A determinant is a sum of products of entries, one from each row. If only one row depends on xx⁠, the determinant is a linear combination of that row's entries with constant coefficients, so differentiating that row is enough.
  2. More generally, if several rows depend on xx⁠, applying the product rule to each product in the full expansion gives the derivative as a sum of determinants, each with one row differentiated (not needed here).

Every fact used, and where it comes from

FactUsed inSource
Expansion of a 3×33\times3 determinant along R1R_1Steps 1, 4Chapter 4, Section 4.2.3
2×22\times2 determinantStep 1Chapter 4, Section 4.2.2
CofactorsStep 2Chapter 4, Section 4.4
Sum and constant multiple rulesStep 3Rules (1) and (2), Section 5.3 (Basics: derivative rules)
Numerical valuesCheckRecomputed in the site's verification script
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