Did you know? You can find the first four terms of a Maclaurin series even when the question never gives y explicitly as a function of x. For example, suppose y satisfies y=1+xy² with y(0)=1. Repeated implicit differentiation gives y′(0)=1, y″(0)=4 and y‴(0)=30. Therefore near zero, y=1+x+2x²+5x³+···. The coefficient five is not guessed: it follows from the chain rule and the stated branch of the implicit equation.
This JC2 H2 Mathematics Tutor Bukit Timah guide answers a narrow A-Level 9758 question: how do you derive the first few Maclaurin coefficients from an implicit equation without first solving explicitly for y? It starts with the exact derivative method, checks it by substitution of a polynomial series, and includes 28 original parameter variations. Each variation is fully auditable through the coefficients it produces and a near-zero convergence check.
The official Singapore–Cambridge H2 Mathematics 9758 syllabus (2027) explicitly includes deriving first Maclaurin terms by repeated implicit differentiation. It excludes questions requiring derivation of a general term of a series. The worked method here stops at the cubic power and keeps the definition of the required branch visible. The mathematical examples are invented instructional equations, not official past-year questions or actual learner outcomes.
Quick answer: evaluate the derivatives at zero
If an implicit equation F(x,y)=0 defines a sufficiently differentiable local function y(x) near x=0, and the branch value y(0)=a is supplied, the cubic Maclaurin expansion is y(x)=a+y′(0)x+[y″(0)/2]x²+[y‴(0)/6]x³+O(x⁴). Differentiate the implicit equation with respect to x, treating y as a function of x each time. Substitute x=0 and the given y(0) only after differentiating correctly. Repeat to obtain the required derivatives.
- Check the initial branch: substitute x=0 and verify the given value y(0) satisfies the original equation.
- Differentiate the y terms: d(y²)/dx=2yy′, not 2y.
- Differentiate products: d(xy²)/dx=y²+2xyy′.
- Evaluate at zero: many x-multiplied terms disappear after differentiating, simplifying the calculation.
- Keep factorials: the coefficient of x² is y″(0)/2!, and the coefficient of x³ is y‴(0)/3!.
- Validate: substitute a truncated series into the original equation and compare the first few coefficients.
First complete worked question: y=1+xy²
Start with the implicit equation y=1+xy² and local branch y(0)=1. Differentiate: y′=y²+2xyy′. Setting x=0 and y=1 gives y′(0)=1. Rewriting as (1−2xy)y′=y² may help isolate the derivative, but the product rule must still be used whenever that equation is differentiated again.
Differentiate y′=y²+2xyy′ once more. The derivative of y² is 2yy′. The derivative of 2xyy′ is 2yy′+2x[(y′)²+yy″]. Therefore y″=4yy′+2x[(y′)²+yy″]. At x=0, y=1 and y′=1, the x-multiplied bracket vanishes, giving y″(0)=4. The quadratic Maclaurin coefficient is 4/2=2.
Differentiate the second-derivative relation again: y‴=6[(y′)²+yy″]+2x[3y′y″+yy‴]. At x=0, the last bracket vanishes. Using y′(0)=1 and y″(0)=4 gives y‴(0)=6(1+4)=30. The cubic coefficient is 30/6=5. Hence the required first four terms are 1+x+2x²+5x³, with a remainder of order x⁴ near zero.
A second route: compare coefficients in the original equation
Suppose we propose y=1+c₁x+c₂x²+c₃x³+O(x⁴). Then y²=1+2c₁x+(c₁²+2c₂)x²+O(x³). Multiply by x and add one: 1+xy²=1+x+2c₁x²+(c₁²+2c₂)x³+O(x⁴). Comparing this with the proposed y series gives c₁=1, c₂=2c₁=2 and c₃=c₁²+2c₂=1+4=5.
This confirms the derivative method without having to compute any more derivatives. It also shows why the x³ coefficient contains contributions from lower powers of y: squaring the series generates cross-terms. If a student’s derivative work gives a different cubic number, coefficient matching identifies which earlier term is feeding the error into the third coefficient.
Why y(0) matters to the series
An implicit equation may describe several local branches. Consider y+y²=x. At x=0, the equation allows y(0)=0 or y(0)=−1. For the branch y(0)=0, coefficient matching gives y=x−x²+2x³+O(x⁴). For the branch y(0)=−1, the other local solution begins y=−1−x+x²−2x³+O(x⁴). The two are not interchangeable.
Both series satisfy the same polynomial relation to the displayed order, but they represent distinct functions near zero. This is why a problem’s initial condition or branch selection is essential. Do not replace it with whichever root seems more convenient after solving the algebra. In a correct implicit Maclaurin solution, the constant term must match y(0) before any derivative calculation begins.
The local implicit-function check
A derivative formula derived by dividing through a coefficient must be examined at the expansion point. For y=1+xy², differentiating and rearranging gives (1−2xy)y′=y². At x=0,y=1 the multiplying factor is 1, so y′ is well defined locally, and repeated differentiation is legitimate on the analytic branch near zero. An equation where the coefficient of y′ vanished at the starting point might require a different analysis or fail to define a unique differentiable branch.
A power series is a local description. It may be accurate near x=0 but need not be valid arbitrarily far away. For the family y=a+bxy² used below with a and b nonzero, solving the quadratic yields a square-root expression, and the Maclaurin expansion on the branch y(0)=a has a natural convergence restriction |4abx|<1. This is the common safe open interval we use for the illustrative calculations.
A numerical check on the original function
For y=1+xy², the local solution passing through y(0)=1 can be written for small nonzero x as y=[1−√(1−4x)]/(2x), with the removable value y(0)=1. At x=0.05, the cubic polynomial predicts 1+0.05+2(0.05²)+5(0.05³)=1.055625. The exact branch is approximately 1.05572809. The small difference is consistent with higher positive powers that were intentionally omitted.
The exact expression is not necessary to derive the first four terms, but it makes a useful independent accuracy check. The approximation deteriorates as x approaches the edge of its convergence interval, particularly near the square-root branch point x=1/4. A learner should not claim the truncated polynomial is exactly equal to the implicit solution for all x.
A general parameter variation: y=a+bxy²
Suppose a and b are fixed nonzero real constants and y(0)=a. Repeated implicit differentiation or coefficient matching gives y=a+(ba²)x+(2b²a³)x²+(5b³a⁴)x³+O(x⁴). The same constant, linear, quadratic and cubic pattern appears regardless of the chosen a and b; only the numerical coefficients change. This formula is a compact verification shortcut for the clinic family, not a derivation of a general nth Maclaurin coefficient.
For example, if a=2 and b=−1, the coefficients are 2, −4, 16 and −80. Thus the local series is y=2−4x+16x²−80x³+O(x⁴). The negative sign on b reverses the odd-power coefficient signs but not the even x² coefficient. This is an accessible way to test whether the sign in a repeated product-rule calculation has been handled correctly.
Differentiate first, then substitute the initial values
When working implicitly, it is often tempting to substitute x=0 and y=a immediately and simplify the original equation into a statement like a=a. That statement contains no information about the derivatives. Instead, differentiate the full x-dependent relation first, then evaluate its derivative at the initial point. Repeat this order for the higher derivatives. This ensures the terms that vanish at x=0 have first contributed correctly to the derivatives.
It is equally important to identify y as a function of x. When differentiating xy², the x contributes y² by the product rule, while y² contributes 2xyy′. Omitting either part produces a wrong first derivative that contaminates every later coefficient. For the worked example the first derivative at zero is one, an easy quick audit before any longer expansion.
Why the cubic term is especially vulnerable
The cubic coefficient depends on both the first and second derivatives. In the coefficient route, it contains c₁² and 2ac₂ terms; these have different algebraic origins. A student who keeps only the square of the linear term or only the cross-term with the quadratic term loses part of the result. This is closely related to the earlier Maclaurin Products: The Cross-Term Trap guide, but here the function is defined implicitly rather than as a displayed product of known series.
After deriving an answer, substitute it into the original equation and compare every power through x³. If a nonzero residual coefficient appears before x⁴, then the claimed Maclaurin polynomial is wrong. This kind of substitution check is more reliable than simply repeating the same derivative calculation that may contain the original error.
An exam-ready sequence for implicit series
- Write the relation F(x,y)=0 and confirm the supplied y(0).
- Differentiate with respect to x, using y′ for derivatives of y.
- Evaluate at x=0 and solve for y′(0).
- Differentiate again, including product and chain terms, and find y″(0).
- Differentiate a third time where the cubic term is required and find y‴(0).
- Write the polynomial with 1/2! and 1/3! factors; use coefficient matching as an independent check.
One-minute Bukit Timah diagnostic
Give a learner y=1+xy², y(0)=1. Ask for y′(0) without a full derivative sheet: one. Then ask for the x² coefficient: two. Then ask for the x³ coefficient: five. Change the equation to y=2−xy², y(0)=2, and ask whether the linear coefficient is still positive. It is −4, because b=−1 and a=2. The pattern tests chain rule, factorial denominators and the branch value together.
Twenty-eight original implicit Maclaurin clinics
The next 28 cases vary the initial branch value a and coefficient b in the implicit equation y=a+bxy². Every case checks coefficients through cubic order and an appropriate near-zero expansion range. The exercises are original mathematical constructions. A student should derive the general coefficient pattern once, then independently verify a handful of parameter choices by differentiating the original equation before accepting a worked line.
1. y=1 − 5xy², y(0)=1
Question. The function y(x) is defined near zero by y=1 − 5xy², with y(0)=1. Derive the first four terms of its Maclaurin series. First verify the initial value, then differentiate implicitly and evaluate the derivatives at x=0. The function is not given explicitly, so do not assume you may choose a different algebraic branch with another initial value.
Complete derivative calculation. With a=1 and b=-5, the first derivative at zero is y′(0)=ba²=-5. Differentiating again gives y″(0)=4b²a³=100. The third derivative at zero is y‴(0)=30b³a⁴=-3750. Dividing by factorials yields the first four terms y=1 − 5x + 50x² − 625x³ + O(x⁴). Each coefficient is exact for the local branch.
Independent check and range. The coefficient-matching route gives c₁=-5, c₂=2bac₁=50, and c₃=b(c₁²+2ac₂)=-625. A safe near-zero expansion range from the explicit square-root branch is |x|<1/20. At x≈0.005, the exact branch evaluates to about 0.976177 and the cubic approximation to 0.9761719. The sign of the x coefficient follows b, while the x² coefficient is positive when a is positive and b is nonzero. This numerical comparison is a check of local behaviour, not a claim that the cubic polynomial equals y everywhere.
2. y=2 − 2xy², y(0)=2
Question. The function y(x) is defined near zero by y=2 − 2xy², with y(0)=2. Derive the first four terms of its Maclaurin series. First verify the initial value, then differentiate implicitly and evaluate the derivatives at x=0. The function is not given explicitly, so do not assume you may choose a different algebraic branch with another initial value.
Complete derivative calculation. With a=2 and b=-2, the first derivative at zero is y′(0)=ba²=-8. Differentiating again gives y″(0)=4b²a³=128. The third derivative at zero is y‴(0)=30b³a⁴=-3840. Dividing by factorials yields the first four terms y=2 − 8x + 64x² − 640x³ + O(x⁴). Each coefficient is exact for the local branch.
Independent check and range. The coefficient-matching route gives c₁=-8, c₂=2bac₁=64, and c₃=b(c₁²+2ac₂)=-640. A safe near-zero expansion range from the explicit square-root branch is |x|<1/16. At x≈0.00625, the exact branch evaluates to about 1.9523539 and the cubic approximation to 1.9523437. The cubic coefficient carries b³, so reversing b reverses the sign of every odd-power coefficient. This numerical comparison is a check of local behaviour, not a claim that the cubic polynomial equals y everywhere.
3. y=3 + 2xy², y(0)=3
Question. The function y(x) is defined near zero by y=3 + 2xy², with y(0)=3. Derive the first four terms of its Maclaurin series. First verify the initial value, then differentiate implicitly and evaluate the derivatives at x=0. The function is not given explicitly, so do not assume you may choose a different algebraic branch with another initial value.
Complete derivative calculation. With a=3 and b=2, the first derivative at zero is y′(0)=ba²=18. Differentiating again gives y″(0)=4b²a³=432. The third derivative at zero is y‴(0)=30b³a⁴=19440. Dividing by factorials yields the first four terms y=3 + 18x + 216x² + 3240x³ + O(x⁴). Each coefficient is exact for the local branch.
Independent check and range. The coefficient-matching route gives c₁=18, c₂=2bac₁=216, and c₃=b(c₁²+2ac₂)=3240. A safe near-zero expansion range from the explicit square-root branch is |x|<1/24. At x≈0.0041667, the exact branch evaluates to about 3.0790021 and the cubic approximation to 3.0789844. The initial value a belongs in the constant term and affects all subsequent coefficients. This numerical comparison is a check of local behaviour, not a claim that the cubic polynomial equals y everywhere.
4. y=4 + 5xy², y(0)=4
Question. The function y(x) is defined near zero by y=4 + 5xy², with y(0)=4. Derive the first four terms of its Maclaurin series. First verify the initial value, then differentiate implicitly and evaluate the derivatives at x=0. The function is not given explicitly, so do not assume you may choose a different algebraic branch with another initial value.
Complete derivative calculation. With a=4 and b=5, the first derivative at zero is y′(0)=ba²=80. Differentiating again gives y″(0)=4b²a³=6400. The third derivative at zero is y‴(0)=30b³a⁴=960000. Dividing by factorials yields the first four terms y=4 + 80x + 3200x² + 160000x³ + O(x⁴). Each coefficient is exact for the local branch.
Independent check and range. The coefficient-matching route gives c₁=80, c₂=2bac₁=3200, and c₃=b(c₁²+2ac₂)=160000. A safe near-zero expansion range from the explicit square-root branch is |x|<1/80. At x≈0.00125, the exact branch evaluates to about 4.1053362 and the cubic approximation to 4.1053125. The factorial denominator in the Maclaurin formula matters: the cubic coefficient is y‴(0)/6. This numerical comparison is a check of local behaviour, not a claim that the cubic polynomial equals y everywhere.
5. y=5 − 3xy², y(0)=5
Question. The function y(x) is defined near zero by y=5 − 3xy², with y(0)=5. Derive the first four terms of its Maclaurin series. First verify the initial value, then differentiate implicitly and evaluate the derivatives at x=0. The function is not given explicitly, so do not assume you may choose a different algebraic branch with another initial value.
Complete derivative calculation. With a=5 and b=-3, the first derivative at zero is y′(0)=ba²=-75. Differentiating again gives y″(0)=4b²a³=4500. The third derivative at zero is y‴(0)=30b³a⁴=-506250. Dividing by factorials yields the first four terms y=5 − 75x + 2250x² − 84375x³ + O(x⁴). Each coefficient is exact for the local branch.
Independent check and range. The coefficient-matching route gives c₁=-75, c₂=2bac₁=2250, and c₃=b(c₁²+2ac₂)=-84375. A safe near-zero expansion range from the explicit square-root branch is |x|<1/60. At x≈0.0016667, the exact branch evaluates to about 4.8808848 and the cubic approximation to 4.8808594. Coefficient matching checks the derivative result without needing an explicit quadratic formula for y(x). This numerical comparison is a check of local behaviour, not a claim that the cubic polynomial equals y everywhere.
6. y=6 + 1xy², y(0)=6
Question. The function y(x) is defined near zero by y=6 + 1xy², with y(0)=6. Derive the first four terms of its Maclaurin series. First verify the initial value, then differentiate implicitly and evaluate the derivatives at x=0. The function is not given explicitly, so do not assume you may choose a different algebraic branch with another initial value.
Complete derivative calculation. With a=6 and b=1, the first derivative at zero is y′(0)=ba²=36. Differentiating again gives y″(0)=4b²a³=864. The third derivative at zero is y‴(0)=30b³a⁴=38880. Dividing by factorials yields the first four terms y=6 + 36x + 432x² + 6480x³ + O(x⁴). Each coefficient is exact for the local branch.
Independent check and range. The coefficient-matching route gives c₁=36, c₂=2bac₁=432, and c₃=b(c₁²+2ac₂)=6480. A safe near-zero expansion range from the explicit square-root branch is |x|<1/24. At x≈0.0041667, the exact branch evaluates to about 6.1580042 and the cubic approximation to 6.1579688. The local power series is a small-x approximation; the stated convergence restriction matters. This numerical comparison is a check of local behaviour, not a claim that the cubic polynomial equals y everywhere.
7. y=2 + 4xy², y(0)=2
Question. The function y(x) is defined near zero by y=2 + 4xy², with y(0)=2. Derive the first four terms of its Maclaurin series. First verify the initial value, then differentiate implicitly and evaluate the derivatives at x=0. The function is not given explicitly, so do not assume you may choose a different algebraic branch with another initial value.
Complete derivative calculation. With a=2 and b=4, the first derivative at zero is y′(0)=ba²=16. Differentiating again gives y″(0)=4b²a³=512. The third derivative at zero is y‴(0)=30b³a⁴=30720. Dividing by factorials yields the first four terms y=2 + 16x + 256x² + 5120x³ + O(x⁴). Each coefficient is exact for the local branch.
Independent check and range. The coefficient-matching route gives c₁=16, c₂=2bac₁=256, and c₃=b(c₁²+2ac₂)=5120. A safe near-zero expansion range from the explicit square-root branch is |x|<1/32. At x≈0.003125, the exact branch evaluates to about 2.0526681 and the cubic approximation to 2.0526562. An expression derived from the implicit equation must be checked on the branch with y(0)=a. This numerical comparison is a check of local behaviour, not a claim that the cubic polynomial equals y everywhere.
8. y=1 − 3xy², y(0)=1
Question. The function y(x) is defined near zero by y=1 − 3xy², with y(0)=1. Derive the first four terms of its Maclaurin series. First verify the initial value, then differentiate implicitly and evaluate the derivatives at x=0. The function is not given explicitly, so do not assume you may choose a different algebraic branch with another initial value.
Complete derivative calculation. With a=1 and b=-3, the first derivative at zero is y′(0)=ba²=-3. Differentiating again gives y″(0)=4b²a³=36. The third derivative at zero is y‴(0)=30b³a⁴=-810. Dividing by factorials yields the first four terms y=1 − 3x + 18x² − 135x³ + O(x⁴). Each coefficient is exact for the local branch.
Independent check and range. The coefficient-matching route gives c₁=-3, c₂=2bac₁=18, and c₃=b(c₁²+2ac₂)=-135. A safe near-zero expansion range from the explicit square-root branch is |x|<1/12. At x≈0.0083333, the exact branch evaluates to about 0.976177 and the cubic approximation to 0.9761719. Do not derive a general nth term for this H2 problem; first four coefficients are sufficient. This numerical comparison is a check of local behaviour, not a claim that the cubic polynomial equals y everywhere.
9. y=2 + 1xy², y(0)=2
Question. The function y(x) is defined near zero by y=2 + 1xy², with y(0)=2. Derive the first four terms of its Maclaurin series. First verify the initial value, then differentiate implicitly and evaluate the derivatives at x=0. The function is not given explicitly, so do not assume you may choose a different algebraic branch with another initial value.
Complete derivative calculation. With a=2 and b=1, the first derivative at zero is y′(0)=ba²=4. Differentiating again gives y″(0)=4b²a³=32. The third derivative at zero is y‴(0)=30b³a⁴=480. Dividing by factorials yields the first four terms y=2 + 4x + 16x² + 80x³ + O(x⁴). Each coefficient is exact for the local branch.
Independent check and range. The coefficient-matching route gives c₁=4, c₂=2bac₁=16, and c₃=b(c₁²+2ac₂)=80. A safe near-zero expansion range from the explicit square-root branch is |x|<1/8. At x≈0.0125, the exact branch evaluates to about 2.0526681 and the cubic approximation to 2.0526562. The sign of the x coefficient follows b, while the x² coefficient is positive when a is positive and b is nonzero. This numerical comparison is a check of local behaviour, not a claim that the cubic polynomial equals y everywhere.
10. y=3 + 4xy², y(0)=3
Question. The function y(x) is defined near zero by y=3 + 4xy², with y(0)=3. Derive the first four terms of its Maclaurin series. First verify the initial value, then differentiate implicitly and evaluate the derivatives at x=0. The function is not given explicitly, so do not assume you may choose a different algebraic branch with another initial value.
Complete derivative calculation. With a=3 and b=4, the first derivative at zero is y′(0)=ba²=36. Differentiating again gives y″(0)=4b²a³=1728. The third derivative at zero is y‴(0)=30b³a⁴=155520. Dividing by factorials yields the first four terms y=3 + 36x + 864x² + 25920x³ + O(x⁴). Each coefficient is exact for the local branch.
Independent check and range. The coefficient-matching route gives c₁=36, c₂=2bac₁=864, and c₃=b(c₁²+2ac₂)=25920. A safe near-zero expansion range from the explicit square-root branch is |x|<1/48. At x≈0.0020833, the exact branch evaluates to about 3.0790021 and the cubic approximation to 3.0789844. The cubic coefficient carries b³, so reversing b reverses the sign of every odd-power coefficient. This numerical comparison is a check of local behaviour, not a claim that the cubic polynomial equals y everywhere.
11. y=4 − 4xy², y(0)=4
Question. The function y(x) is defined near zero by y=4 − 4xy², with y(0)=4. Derive the first four terms of its Maclaurin series. First verify the initial value, then differentiate implicitly and evaluate the derivatives at x=0. The function is not given explicitly, so do not assume you may choose a different algebraic branch with another initial value.
Complete derivative calculation. With a=4 and b=-4, the first derivative at zero is y′(0)=ba²=-64. Differentiating again gives y″(0)=4b²a³=4096. The third derivative at zero is y‴(0)=30b³a⁴=-491520. Dividing by factorials yields the first four terms y=4 − 64x + 2048x² − 81920x³ + O(x⁴). Each coefficient is exact for the local branch.
Independent check and range. The coefficient-matching route gives c₁=-64, c₂=2bac₁=2048, and c₃=b(c₁²+2ac₂)=-81920. A safe near-zero expansion range from the explicit square-root branch is |x|<1/64. At x≈0.0015625, the exact branch evaluates to about 3.9047079 and the cubic approximation to 3.9046875. The initial value a belongs in the constant term and affects all subsequent coefficients. This numerical comparison is a check of local behaviour, not a claim that the cubic polynomial equals y everywhere.
12. y=5 − 1xy², y(0)=5
Question. The function y(x) is defined near zero by y=5 − 1xy², with y(0)=5. Derive the first four terms of its Maclaurin series. First verify the initial value, then differentiate implicitly and evaluate the derivatives at x=0. The function is not given explicitly, so do not assume you may choose a different algebraic branch with another initial value.
Complete derivative calculation. With a=5 and b=-1, the first derivative at zero is y′(0)=ba²=-25. Differentiating again gives y″(0)=4b²a³=500. The third derivative at zero is y‴(0)=30b³a⁴=-18750. Dividing by factorials yields the first four terms y=5 − 25x + 250x² − 3125x³ + O(x⁴). Each coefficient is exact for the local branch.
Independent check and range. The coefficient-matching route gives c₁=-25, c₂=2bac₁=250, and c₃=b(c₁²+2ac₂)=-3125. A safe near-zero expansion range from the explicit square-root branch is |x|<1/20. At x≈0.005, the exact branch evaluates to about 4.8808848 and the cubic approximation to 4.8808594. The factorial denominator in the Maclaurin formula matters: the cubic coefficient is y‴(0)/6. This numerical comparison is a check of local behaviour, not a claim that the cubic polynomial equals y everywhere.
13. y=6 + 3xy², y(0)=6
Question. The function y(x) is defined near zero by y=6 + 3xy², with y(0)=6. Derive the first four terms of its Maclaurin series. First verify the initial value, then differentiate implicitly and evaluate the derivatives at x=0. The function is not given explicitly, so do not assume you may choose a different algebraic branch with another initial value.
Complete derivative calculation. With a=6 and b=3, the first derivative at zero is y′(0)=ba²=108. Differentiating again gives y″(0)=4b²a³=7776. The third derivative at zero is y‴(0)=30b³a⁴=1049760. Dividing by factorials yields the first four terms y=6 + 108x + 3888x² + 174960x³ + O(x⁴). Each coefficient is exact for the local branch.
Independent check and range. The coefficient-matching route gives c₁=108, c₂=2bac₁=3888, and c₃=b(c₁²+2ac₂)=174960. A safe near-zero expansion range from the explicit square-root branch is |x|<1/72. At x≈0.0013889, the exact branch evaluates to about 6.1580042 and the cubic approximation to 6.1579688. Coefficient matching checks the derivative result without needing an explicit quadratic formula for y(x). This numerical comparison is a check of local behaviour, not a claim that the cubic polynomial equals y everywhere.
14. y=2 − 5xy², y(0)=2
Question. The function y(x) is defined near zero by y=2 − 5xy², with y(0)=2. Derive the first four terms of its Maclaurin series. First verify the initial value, then differentiate implicitly and evaluate the derivatives at x=0. The function is not given explicitly, so do not assume you may choose a different algebraic branch with another initial value.
Complete derivative calculation. With a=2 and b=-5, the first derivative at zero is y′(0)=ba²=-20. Differentiating again gives y″(0)=4b²a³=800. The third derivative at zero is y‴(0)=30b³a⁴=-60000. Dividing by factorials yields the first four terms y=2 − 20x + 400x² − 10000x³ + O(x⁴). Each coefficient is exact for the local branch.
Independent check and range. The coefficient-matching route gives c₁=-20, c₂=2bac₁=400, and c₃=b(c₁²+2ac₂)=-10000. A safe near-zero expansion range from the explicit square-root branch is |x|<1/40. At x≈0.0025, the exact branch evaluates to about 1.9523539 and the cubic approximation to 1.9523437. The local power series is a small-x approximation; the stated convergence restriction matters. This numerical comparison is a check of local behaviour, not a claim that the cubic polynomial equals y everywhere.
15. y=1 − 1xy², y(0)=1
Question. The function y(x) is defined near zero by y=1 − 1xy², with y(0)=1. Derive the first four terms of its Maclaurin series. First verify the initial value, then differentiate implicitly and evaluate the derivatives at x=0. The function is not given explicitly, so do not assume you may choose a different algebraic branch with another initial value.
Complete derivative calculation. With a=1 and b=-1, the first derivative at zero is y′(0)=ba²=-1. Differentiating again gives y″(0)=4b²a³=4. The third derivative at zero is y‴(0)=30b³a⁴=-30. Dividing by factorials yields the first four terms y=1 − 1x + 2x² − 5x³ + O(x⁴). Each coefficient is exact for the local branch.
Independent check and range. The coefficient-matching route gives c₁=-1, c₂=2bac₁=2, and c₃=b(c₁²+2ac₂)=-5. A safe near-zero expansion range from the explicit square-root branch is |x|<1/4. At x≈0.025, the exact branch evaluates to about 0.976177 and the cubic approximation to 0.9761719. An expression derived from the implicit equation must be checked on the branch with y(0)=a. This numerical comparison is a check of local behaviour, not a claim that the cubic polynomial equals y everywhere.
16. y=2 + 3xy², y(0)=2
Question. The function y(x) is defined near zero by y=2 + 3xy², with y(0)=2. Derive the first four terms of its Maclaurin series. First verify the initial value, then differentiate implicitly and evaluate the derivatives at x=0. The function is not given explicitly, so do not assume you may choose a different algebraic branch with another initial value.
Complete derivative calculation. With a=2 and b=3, the first derivative at zero is y′(0)=ba²=12. Differentiating again gives y″(0)=4b²a³=288. The third derivative at zero is y‴(0)=30b³a⁴=12960. Dividing by factorials yields the first four terms y=2 + 12x + 144x² + 2160x³ + O(x⁴). Each coefficient is exact for the local branch.
Independent check and range. The coefficient-matching route gives c₁=12, c₂=2bac₁=144, and c₃=b(c₁²+2ac₂)=2160. A safe near-zero expansion range from the explicit square-root branch is |x|<1/24. At x≈0.0041667, the exact branch evaluates to about 2.0526681 and the cubic approximation to 2.0526562. Do not derive a general nth term for this H2 problem; first four coefficients are sufficient. This numerical comparison is a check of local behaviour, not a claim that the cubic polynomial equals y everywhere.
17. y=3 − 5xy², y(0)=3
Question. The function y(x) is defined near zero by y=3 − 5xy², with y(0)=3. Derive the first four terms of its Maclaurin series. First verify the initial value, then differentiate implicitly and evaluate the derivatives at x=0. The function is not given explicitly, so do not assume you may choose a different algebraic branch with another initial value.
Complete derivative calculation. With a=3 and b=-5, the first derivative at zero is y′(0)=ba²=-45. Differentiating again gives y″(0)=4b²a³=2700. The third derivative at zero is y‴(0)=30b³a⁴=-303750. Dividing by factorials yields the first four terms y=3 − 45x + 1350x² − 50625x³ + O(x⁴). Each coefficient is exact for the local branch.
Independent check and range. The coefficient-matching route gives c₁=-45, c₂=2bac₁=1350, and c₃=b(c₁²+2ac₂)=-50625. A safe near-zero expansion range from the explicit square-root branch is |x|<1/60. At x≈0.0016667, the exact branch evaluates to about 2.9285309 and the cubic approximation to 2.9285156. The sign of the x coefficient follows b, while the x² coefficient is positive when a is positive and b is nonzero. This numerical comparison is a check of local behaviour, not a claim that the cubic polynomial equals y everywhere.
18. y=4 − 2xy², y(0)=4
Question. The function y(x) is defined near zero by y=4 − 2xy², with y(0)=4. Derive the first four terms of its Maclaurin series. First verify the initial value, then differentiate implicitly and evaluate the derivatives at x=0. The function is not given explicitly, so do not assume you may choose a different algebraic branch with another initial value.
Complete derivative calculation. With a=4 and b=-2, the first derivative at zero is y′(0)=ba²=-32. Differentiating again gives y″(0)=4b²a³=1024. The third derivative at zero is y‴(0)=30b³a⁴=-61440. Dividing by factorials yields the first four terms y=4 − 32x + 512x² − 10240x³ + O(x⁴). Each coefficient is exact for the local branch.
Independent check and range. The coefficient-matching route gives c₁=-32, c₂=2bac₁=512, and c₃=b(c₁²+2ac₂)=-10240. A safe near-zero expansion range from the explicit square-root branch is |x|<1/32. At x≈0.003125, the exact branch evaluates to about 3.9047079 and the cubic approximation to 3.9046875. The cubic coefficient carries b³, so reversing b reverses the sign of every odd-power coefficient. This numerical comparison is a check of local behaviour, not a claim that the cubic polynomial equals y everywhere.
19. y=5 + 2xy², y(0)=5
Question. The function y(x) is defined near zero by y=5 + 2xy², with y(0)=5. Derive the first four terms of its Maclaurin series. First verify the initial value, then differentiate implicitly and evaluate the derivatives at x=0. The function is not given explicitly, so do not assume you may choose a different algebraic branch with another initial value.
Complete derivative calculation. With a=5 and b=2, the first derivative at zero is y′(0)=ba²=50. Differentiating again gives y″(0)=4b²a³=2000. The third derivative at zero is y‴(0)=30b³a⁴=150000. Dividing by factorials yields the first four terms y=5 + 50x + 1000x² + 25000x³ + O(x⁴). Each coefficient is exact for the local branch.
Independent check and range. The coefficient-matching route gives c₁=50, c₂=2bac₁=1000, and c₃=b(c₁²+2ac₂)=25000. A safe near-zero expansion range from the explicit square-root branch is |x|<1/40. At x≈0.0025, the exact branch evaluates to about 5.1316702 and the cubic approximation to 5.1316406. The initial value a belongs in the constant term and affects all subsequent coefficients. This numerical comparison is a check of local behaviour, not a claim that the cubic polynomial equals y everywhere.
20. y=6 + 5xy², y(0)=6
Question. The function y(x) is defined near zero by y=6 + 5xy², with y(0)=6. Derive the first four terms of its Maclaurin series. First verify the initial value, then differentiate implicitly and evaluate the derivatives at x=0. The function is not given explicitly, so do not assume you may choose a different algebraic branch with another initial value.
Complete derivative calculation. With a=6 and b=5, the first derivative at zero is y′(0)=ba²=180. Differentiating again gives y″(0)=4b²a³=21600. The third derivative at zero is y‴(0)=30b³a⁴=4860000. Dividing by factorials yields the first four terms y=6 + 180x + 10800x² + 810000x³ + O(x⁴). Each coefficient is exact for the local branch.
Independent check and range. The coefficient-matching route gives c₁=180, c₂=2bac₁=10800, and c₃=b(c₁²+2ac₂)=810000. A safe near-zero expansion range from the explicit square-root branch is |x|<1/120. At x≈0.0008333, the exact branch evaluates to about 6.1580042 and the cubic approximation to 6.1579688. The factorial denominator in the Maclaurin formula matters: the cubic coefficient is y‴(0)/6. This numerical comparison is a check of local behaviour, not a claim that the cubic polynomial equals y everywhere.
21. y=2 − 3xy², y(0)=2
Question. The function y(x) is defined near zero by y=2 − 3xy², with y(0)=2. Derive the first four terms of its Maclaurin series. First verify the initial value, then differentiate implicitly and evaluate the derivatives at x=0. The function is not given explicitly, so do not assume you may choose a different algebraic branch with another initial value.
Complete derivative calculation. With a=2 and b=-3, the first derivative at zero is y′(0)=ba²=-12. Differentiating again gives y″(0)=4b²a³=288. The third derivative at zero is y‴(0)=30b³a⁴=-12960. Dividing by factorials yields the first four terms y=2 − 12x + 144x² − 2160x³ + O(x⁴). Each coefficient is exact for the local branch.
Independent check and range. The coefficient-matching route gives c₁=-12, c₂=2bac₁=144, and c₃=b(c₁²+2ac₂)=-2160. A safe near-zero expansion range from the explicit square-root branch is |x|<1/24. At x≈0.0041667, the exact branch evaluates to about 1.9523539 and the cubic approximation to 1.9523437. Coefficient matching checks the derivative result without needing an explicit quadratic formula for y(x). This numerical comparison is a check of local behaviour, not a claim that the cubic polynomial equals y everywhere.
22. y=1 + 2xy², y(0)=1
Question. The function y(x) is defined near zero by y=1 + 2xy², with y(0)=1. Derive the first four terms of its Maclaurin series. First verify the initial value, then differentiate implicitly and evaluate the derivatives at x=0. The function is not given explicitly, so do not assume you may choose a different algebraic branch with another initial value.
Complete derivative calculation. With a=1 and b=2, the first derivative at zero is y′(0)=ba²=2. Differentiating again gives y″(0)=4b²a³=16. The third derivative at zero is y‴(0)=30b³a⁴=240. Dividing by factorials yields the first four terms y=1 + 2x + 8x² + 40x³ + O(x⁴). Each coefficient is exact for the local branch.
Independent check and range. The coefficient-matching route gives c₁=2, c₂=2bac₁=8, and c₃=b(c₁²+2ac₂)=40. A safe near-zero expansion range from the explicit square-root branch is |x|<1/8. At x≈0.0125, the exact branch evaluates to about 1.026334 and the cubic approximation to 1.0263281. The local power series is a small-x approximation; the stated convergence restriction matters. This numerical comparison is a check of local behaviour, not a claim that the cubic polynomial equals y everywhere.
23. y=2 + 5xy², y(0)=2
Question. The function y(x) is defined near zero by y=2 + 5xy², with y(0)=2. Derive the first four terms of its Maclaurin series. First verify the initial value, then differentiate implicitly and evaluate the derivatives at x=0. The function is not given explicitly, so do not assume you may choose a different algebraic branch with another initial value.
Complete derivative calculation. With a=2 and b=5, the first derivative at zero is y′(0)=ba²=20. Differentiating again gives y″(0)=4b²a³=800. The third derivative at zero is y‴(0)=30b³a⁴=60000. Dividing by factorials yields the first four terms y=2 + 20x + 400x² + 10000x³ + O(x⁴). Each coefficient is exact for the local branch.
Independent check and range. The coefficient-matching route gives c₁=20, c₂=2bac₁=400, and c₃=b(c₁²+2ac₂)=10000. A safe near-zero expansion range from the explicit square-root branch is |x|<1/40. At x≈0.0025, the exact branch evaluates to about 2.0526681 and the cubic approximation to 2.0526562. An expression derived from the implicit equation must be checked on the branch with y(0)=a. This numerical comparison is a check of local behaviour, not a claim that the cubic polynomial equals y everywhere.
24. y=3 − 3xy², y(0)=3
Question. The function y(x) is defined near zero by y=3 − 3xy², with y(0)=3. Derive the first four terms of its Maclaurin series. First verify the initial value, then differentiate implicitly and evaluate the derivatives at x=0. The function is not given explicitly, so do not assume you may choose a different algebraic branch with another initial value.
Complete derivative calculation. With a=3 and b=-3, the first derivative at zero is y′(0)=ba²=-27. Differentiating again gives y″(0)=4b²a³=972. The third derivative at zero is y‴(0)=30b³a⁴=-65610. Dividing by factorials yields the first four terms y=3 − 27x + 486x² − 10935x³ + O(x⁴). Each coefficient is exact for the local branch.
Independent check and range. The coefficient-matching route gives c₁=-27, c₂=2bac₁=486, and c₃=b(c₁²+2ac₂)=-10935. A safe near-zero expansion range from the explicit square-root branch is |x|<1/36. At x≈0.0027778, the exact branch evaluates to about 2.9285309 and the cubic approximation to 2.9285156. Do not derive a general nth term for this H2 problem; first four coefficients are sufficient. This numerical comparison is a check of local behaviour, not a claim that the cubic polynomial equals y everywhere.
25. y=4 + 1xy², y(0)=4
Question. The function y(x) is defined near zero by y=4 + 1xy², with y(0)=4. Derive the first four terms of its Maclaurin series. First verify the initial value, then differentiate implicitly and evaluate the derivatives at x=0. The function is not given explicitly, so do not assume you may choose a different algebraic branch with another initial value.
Complete derivative calculation. With a=4 and b=1, the first derivative at zero is y′(0)=ba²=16. Differentiating again gives y″(0)=4b²a³=256. The third derivative at zero is y‴(0)=30b³a⁴=7680. Dividing by factorials yields the first four terms y=4 + 16x + 128x² + 1280x³ + O(x⁴). Each coefficient is exact for the local branch.
Independent check and range. The coefficient-matching route gives c₁=16, c₂=2bac₁=128, and c₃=b(c₁²+2ac₂)=1280. A safe near-zero expansion range from the explicit square-root branch is |x|<1/16. At x≈0.00625, the exact branch evaluates to about 4.1053362 and the cubic approximation to 4.1053125. The sign of the x coefficient follows b, while the x² coefficient is positive when a is positive and b is nonzero. This numerical comparison is a check of local behaviour, not a claim that the cubic polynomial equals y everywhere.
26. y=5 + 4xy², y(0)=5
Question. The function y(x) is defined near zero by y=5 + 4xy², with y(0)=5. Derive the first four terms of its Maclaurin series. First verify the initial value, then differentiate implicitly and evaluate the derivatives at x=0. The function is not given explicitly, so do not assume you may choose a different algebraic branch with another initial value.
Complete derivative calculation. With a=5 and b=4, the first derivative at zero is y′(0)=ba²=100. Differentiating again gives y″(0)=4b²a³=8000. The third derivative at zero is y‴(0)=30b³a⁴=1200000. Dividing by factorials yields the first four terms y=5 + 100x + 4000x² + 200000x³ + O(x⁴). Each coefficient is exact for the local branch.
Independent check and range. The coefficient-matching route gives c₁=100, c₂=2bac₁=4000, and c₃=b(c₁²+2ac₂)=200000. A safe near-zero expansion range from the explicit square-root branch is |x|<1/80. At x≈0.00125, the exact branch evaluates to about 5.1316702 and the cubic approximation to 5.1316406. The cubic coefficient carries b³, so reversing b reverses the sign of every odd-power coefficient. This numerical comparison is a check of local behaviour, not a claim that the cubic polynomial equals y everywhere.
27. y=6 − 4xy², y(0)=6
Question. The function y(x) is defined near zero by y=6 − 4xy², with y(0)=6. Derive the first four terms of its Maclaurin series. First verify the initial value, then differentiate implicitly and evaluate the derivatives at x=0. The function is not given explicitly, so do not assume you may choose a different algebraic branch with another initial value.
Complete derivative calculation. With a=6 and b=-4, the first derivative at zero is y′(0)=ba²=-144. Differentiating again gives y″(0)=4b²a³=13824. The third derivative at zero is y‴(0)=30b³a⁴=-2488320. Dividing by factorials yields the first four terms y=6 − 144x + 6912x² − 414720x³ + O(x⁴). Each coefficient is exact for the local branch.
Independent check and range. The coefficient-matching route gives c₁=-144, c₂=2bac₁=6912, and c₃=b(c₁²+2ac₂)=-414720. A safe near-zero expansion range from the explicit square-root branch is |x|<1/96. At x≈0.0010417, the exact branch evaluates to about 5.8570618 and the cubic approximation to 5.8570313. The initial value a belongs in the constant term and affects all subsequent coefficients. This numerical comparison is a check of local behaviour, not a claim that the cubic polynomial equals y everywhere.
28. y=2 − 1xy², y(0)=2
Question. The function y(x) is defined near zero by y=2 − 1xy², with y(0)=2. Derive the first four terms of its Maclaurin series. First verify the initial value, then differentiate implicitly and evaluate the derivatives at x=0. The function is not given explicitly, so do not assume you may choose a different algebraic branch with another initial value.
Complete derivative calculation. With a=2 and b=-1, the first derivative at zero is y′(0)=ba²=-4. Differentiating again gives y″(0)=4b²a³=32. The third derivative at zero is y‴(0)=30b³a⁴=-480. Dividing by factorials yields the first four terms y=2 − 4x + 16x² − 80x³ + O(x⁴). Each coefficient is exact for the local branch.
Independent check and range. The coefficient-matching route gives c₁=-4, c₂=2bac₁=16, and c₃=b(c₁²+2ac₂)=-80. A safe near-zero expansion range from the explicit square-root branch is |x|<1/8. At x≈0.0125, the exact branch evaluates to about 1.9523539 and the cubic approximation to 1.9523437. The factorial denominator in the Maclaurin formula matters: the cubic coefficient is y‴(0)/6. This numerical comparison is a check of local behaviour, not a claim that the cubic polynomial equals y everywhere.
Eight short answers for revision
- y=1+xy² with y(0)=1: y=1+x+2x²+5x³+O(x⁴).
- The first three derivatives there are y′(0)=1, y″(0)=4, y‴(0)=30.
- y=2−xy² with y(0)=2: y=2−4x+16x²−80x³+O(x⁴).
- y=1−xy² with y(0)=1: y=1−x+2x²−5x³+O(x⁴).
- y+y²=x, y(0)=0: y=x−x²+2x³+O(x⁴).
- The other branch y+y²=x, y(0)=−1 starts y=−1−x+x²−2x³+O(x⁴).
- A cubic coefficient uses y‴(0)/3!, not y‴(0).
- For y=a+bxy² with nonzero a,b, |4abx|<1 is a safe open range for the stated local series.
Five-day study and independent-check plan
Day one: practise first implicit derivatives and evaluate them at the initial condition. Day two: differentiate the resulting relation again, showing every product-rule term. Day three: obtain cubic derivatives and remember the factorials in the Maclaurin polynomial. Day four: use coefficient matching from the original implicit equation as an independent route. Day five: mix different starting branch values and signed parameters, then check whether the resulting polynomial approximates the defined function near x=0.
In a three-student tutorial, one pupil can derive y′(0), another can work through the second and third derivatives, and the third can audit the result by coefficient matching. Rotate roles and change a or b. The purpose is to build an independent verification route rather than repeating an error in the same symbolic method.
Frequently asked questions
Can I find a Maclaurin series without solving for y explicitly?
Yes. Differentiate the implicit equation repeatedly, evaluate derivatives at the stated point and substitute into the Maclaurin formula.
Why must the question specify y(0)?
An implicit relation can have multiple branches. The stated value identifies which local function and corresponding series is intended.
Where does the 5x³ coefficient come from?
For y=1+xy², implicit differentiation gives y‴(0)=30, and 30/3!=5. Coefficient matching independently produces c₃=1²+2(2)=5.
Does the series equal the function everywhere?
No. A finite truncation is a local polynomial approximation near zero, and the relevant convergence range must be respected.
Why do the derivative formulas contain y′ and y″?
Because y depends on x. The product and chain rules must differentiate that dependence rather than treating y as a constant.
Does H2 require deriving the general nth term?
No. The official 2027 syllabus excludes problems requiring derivation of a general series term, while including the first few terms by repeated implicit differentiation.
What is the easiest independent check?
Insert the cubic polynomial into the original equation and check that coefficients through x³ match on both sides.
Can the choice of initial branch change the series?
Yes. For the relation y+y²=x, the branches through y=0 and y=−1 at x=0 have different constant and higher coefficients.
Related H2 guides and official syllabus
- Circular Seating: When Rotations Count as the Same
- Rational Inequalities: Why Denominator Zeros Stay Open
- Maclaurin by Implicit Differentiation: First Four Terms: current guide
- Recurrence Sequences: Fixed Points vs Actual Terms
See the broader Bukit Timah H2 Maclaurin Series guide, the Quotient Series and Missing Coefficients article, and the JC2 Bukit Timah Tuition hub. The immutable Clementi three-pupil Mathematics tutoring reference explains the small-group format at another level.
Official source: SEAB Singapore–Cambridge H2 Mathematics 9758 2027 syllabus, Maclaurin Series §5.2. All equations and parameter variations are original instructional examples, not official examination questions, records of tutoring results or performance guarantees.