Approximation Using Differentials
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Direct answer
Small changes obey a linear law: for y = f(x), the differential dy = f'(x)·dx approximates the true change Δy when dx is small, so f(x + Δx) ≈ f(x) + f'(x)·Δx. That single line powers every JEE estimate — √26 ≈ 5.1 from √25 = 5 with f'(x) = 1/(2√x) = 1/10 and Δx = 1; (2.002)³ ≈ 8.024 from x³ at x = 2 with slope 3x² = 12. Errors propagate through the same machinery: relative error in a product of measured quantities adds, so a 1% error in a sphere's radius produces a 3% error in volume (dV/V = 3 dr/r). The method's accuracy is first-order — it ignores Δx² terms — which is exactly why the approximations land close but not exact.
What you must remember
- Core identity: Δy ≈ dy = f'(x)dx, and f(x + dx) ≈ f(x) + f'(x)·dx; choose x a perfect base (25, 8, 16...) near the target so f(x) and f'(x) are exact.
- Standard estimates: √(a + h) ≈ √a + h/(2√a); (a + h)^n ≈ a^n + n·a^(n−1)·h; sin θ ≈ θ and ln(1 + h) ≈ h for small h in radians.
- Relative error language: dV/V = 3 dr/r for a sphere (V ∝ r³); dA/A = 2 dr/r for a circle; percentage error is 100 × relative error.
- Product rule for errors: for z = x·y, dz/z = dx/x + dy/y; for z = x/y, the relative errors subtract — relative errors add in magnitude for products and powers.
- Power law: for y = x^n, relative error multiplies by n: dy/y = n·dx/x — one derivative computation covers all such questions.
- Radian discipline: trigonometric approximations sin h ≈ h and tan h ≈ h demand h in radians; degree inputs wreck the estimate silently.
- Accuracy caveat: the error committed is of order (dx)² — fine for two-decimal answers, the reason (1.02)^10 needs binomial terms rather than a single differential if four decimals are wanted.
Estimating √26 and friends
Estimate √26. Set f(x) = √x, take the base point x = 25 where everything is exact: f(25) = 5, f'(25) = 1/(2√25) = 1/10. With dx = 1, √26 ≈ 5 + (1/10)(1) = 5.1; the true value 5.0990... confirms the error is under 0.002%. The pattern repeats everywhere: (0.98)^5? Take f(x) = x⁵ at x = 1: f'(1) = 5, dx = −0.02, estimate 1 + 5(−0.02) = 0.9. A cube edge measured as 3.002 cm: V = x³ at 3 gives 27 + 27(0.002) = 27.054, using slope 3x² = 27.
Error propagation is the same derivative wearing work clothes. The radius of a sphere is measured with 2% error; what is the error in volume? V = (4/3)πr³, so dV/V = 3 dr/r = 6%. In a cylinder with 1% error in radius and 2% in height, V = πr²h gives dV/V = 2(1%) + 2% = 4%. Notice how the exponents become coefficients: differentiation converts every "percentage" question into arithmetic on exponents, and that conversion — not the specific numbers — is the examinable idea.
Order-of-error traps
The recurring slip is choosing a poor base point: approximating √26 from x = 16 forces dx = 10 and a visibly worse estimate; the art is picking the nearest perfect value. The second slip is unit blindness in trigonometric estimates — sin(2°) is 2π/180 ≈ 0.035, not 0.02; the differential h must be radians. Third, sign errors when dx is negative (estimates below the base) invert the correction direction. JEE Main asks direct estimates as numerical answers, typically to two or three decimals, where the linear term suffices; Advanced embeds approximation inside proofs (show that √(1 + x) ≈ 1 + x/2 with error less than x²/8) or asks which measurement dominates a compound error — questions about method, not arithmetic. A final viva-worthy point: the differential approximation is the first Taylor polynomial, so "differentials versus binomial expansion" is really a choice of how many correction terms the required precision demands.
Frequently asked questions
What is the differential approximation formula?
f(x + dx) ≈ f(x) + f'(x)·dx — the function value plus slope times small increment, valid while dx stays small.
How do you estimate √26 without a calculator?
Use √25 = 5 and f'(25) = 1/10: √26 ≈ 5 + 1/10 = 5.1.
A 1% error in a sphere's radius causes what error in its volume?
3%, because V ∝ r³ makes dV/V = 3 dr/r — exponents become percentage multipliers.
How do relative errors combine in a product or quotient?
They add for products (dz/z = dx/x + dy/y) and subtract for quotients; in magnitude, propagated relative errors always accumulate.
Why do trigonometric approximations require radians?
Because sin h ≈ h and tan h ≈ h come from limits defined with h in radians; degree-measured angles must first be converted, or the estimate is scaled incorrectly by π/180.