1. Chapter at a Glance
The derivative measures the rate of change of a quantity and the slope of a curve. This chapter turns that idea into tools: finding tangents and normals, deciding where a function increases or decreases, locating maxima and minima, and approximating values. JEE Main asks one or two questions — often on maxima/minima, monotonicity and tangents.
- Local maximum and minimum values of polynomials and trigonometric functions.
- Intervals where a function is increasing or decreasing.
- Equations of tangents and normals; angle between curves.
- Rolle's theorem and the Mean Value Theorem; optimisation word problems.
2. Rate of Change
If y = f(x), then dy/dx is the rate of change of y with respect to x. When both depend on time t, use the chain rule: dy/dt = (dy/dx)(dx/dt).
- Circle: A = πr² ⇒ dA/dt = 2πr·dr/dt.
- Sphere: V = (4/3)πr³ ⇒ dV/dt = 4πr²·dr/dt.
- Small changes (approximation): Δy ≈ f′(x)Δx; percentage error in y = xⁿ is n times that in x.
3. Tangents and Normals
The slope of the tangent to y = f(x) at (x1, y1) is m = f′(x1).
- Tangent parallel to the x-axis: dy/dx = 0. Tangent parallel to the y-axis: dx/dy = 0.
- Angle between two curves at a common point = angle between their tangents: tan θ = |(m1 − m2)/(1 + m1m2)|. Curves cut orthogonally if m1m2 = −1.
4. Increasing and Decreasing Functions
- f′(x) > 0 on an interval ⇒ f is strictly increasing there.
- f′(x) < 0 on an interval ⇒ f is strictly decreasing there.
- f′(x) ≥ 0 with equality only at isolated points still gives strictly increasing (e.g. x³).
Method: find f′(x), find where f′(x) = 0 or is undefined (critical points), and test the sign of f′ in each interval — a sign chart.
5. Maxima and Minima
First derivative test
At a critical point c: if f′ changes from + to −, f has a local maximum; from − to +, a local minimum; no change, neither (a point of inflection).
Second derivative test
If f′(c) = 0: f″(c) < 0 ⇒ local maximum; f″(c) > 0 ⇒ local minimum; f″(c) = 0 ⇒ test fails, use the first derivative test.
Absolute (global) extremes on [a, b]
Compare the values of f at all critical points inside (a, b) and at the end points a and b. The largest is the absolute maximum, the smallest the absolute minimum.
- Of all rectangles with a given perimeter, the square has the largest area.
- Of all rectangles with a given area, the square has the least perimeter.
- The largest cylinder inscribed in a sphere of radius R has height 2R/√3 and volume 4πR³/(3√3).
- Open box from a square sheet of side a by cutting squares of side x from the corners: volume is greatest when x = a/6.
6. Rolle's Theorem and the Mean Value Theorem
- Rolle's theorem: if f is continuous on [a, b], differentiable on (a, b) and f(a) = f(b), then f′(c) = 0 for at least one c in (a, b).
- Lagrange's Mean Value Theorem (LMVT): if f is continuous on [a, b] and differentiable on (a, b), then for some c in (a, b):
Geometrically: somewhere the tangent is parallel to the chord joining the end points.
7. Formula Master Table
| Tool | Formula / Condition |
|---|---|
| Slope of tangent | m = f′(x1) |
| Tangent | y − y1 = m(x − x1) |
| Normal | y − y1 = −(1/m)(x − x1) |
| Angle between curves | tan θ = |(m1 − m2)/(1 + m1m2)| |
| Increasing / decreasing | f′ > 0 / f′ < 0 |
| Local max / min | f′ = 0 with f″ < 0 / f″ > 0 |
| Approximation | f(x + Δx) ≈ f(x) + f′(x)Δx |
| LMVT | f′(c) = (f(b) − f(a))/(b − a) |
| Length of subtangent / subnormal | |y/y′| / |y y′| |
8. Mind Map
- dy/dt = y′ · dx/dt
- Area, volume problems
- Approximation
- Slope f′(x)
- Normal slope −1/m
- Angle between curves
- Sign chart of f′
- Critical points
- 1st and 2nd derivative tests
- Absolute on [a, b]
- Rolle, LMVT
9. Tips and Tricks
- a sin x + b cos x has maximum √(a² + b²) and minimum −√(a² + b²) — no calculus needed.
- For a cubic, the local maximum and minimum occur at the roots of f′(x) = 0, and their x-values are symmetric about the point of inflection x = −b/3a.
- In word problems, write the quantity to optimise as a function of one variable using the constraint before differentiating.
- For "tangent passes through a given external point", use the point of contact (t, f(t)) as the unknown.
- Forgetting to check the end points when finding absolute extremes on a closed interval.
- Calling every critical point an extremum — x³ has f′(0) = 0 but no extremum at 0.
- Using Rolle's theorem without checking f(a) = f(b) and differentiability (e.g. |x| on [−1, 1]).
10. Important One-liners
- A local maximum value can be smaller than a local minimum value of the same function.
- A continuous function on a closed interval always attains an absolute maximum and minimum.
- If f′(x) > 0 everywhere, f is one-one and so f(x) = k has at most one root.
- The normal to a circle always passes through its centre.
- At a point of inflection, the curve changes from concave up to concave down or vice versa (f″ changes sign).
- Between two roots of f, there is at least one root of f′ (Rolle).
- For y = eˣ, the slope at any point equals the y-coordinate.
11. PYQ-pattern Questions with Solutions
These questions are original and modelled on concepts frequently tested in JEE Main.
Find the local maximum value of f(x) = x³ − 3x² − 9x + 5.
Show solution
f′(x) = 3x² − 6x − 9 = 3(x − 3)(x + 1), so the critical points are x = −1 and x = 3.
f″(x) = 6x − 6: f″(−1) = −12 < 0, so x = −1 is a local maximum. f(−1) = −1 − 3 + 9 + 5 = 10.
(At x = 3, f″ = 12 > 0, a local minimum with value −22.)
Answer: 10
The equation of the tangent to y = x² at the point (1, 1) is:
- y = 2x − 1
- y = x
- y = 2x + 1
- y = −2x + 3
Show solution
Slope = 2x = 2 at x = 1. Tangent: y − 1 = 2(x − 1) ⇒ y = 2x − 1.
Answer: (A)
The radius of a circle increases at 0.5 cm/s. At the moment the radius is 10 cm, the area increases at:
- 5π cm²/s
- 10π cm²/s
- 20π cm²/s
- π cm²/s
Show solution
dA/dt = 2πr·dr/dt = 2π × 10 × 0.5 = 10π cm²/s.
Answer: (B)
The function f(x) = x³ − 3x² − 9x + 5 is strictly decreasing on:
- (−∞, −1)
- (−1, 3)
- (3, ∞)
- (−3, 1)
Show solution
f′(x) = 3(x − 3)(x + 1) is negative between its roots, i.e. for −1 < x < 3.
Answer: (B)
The value of c given by Lagrange's Mean Value Theorem for f(x) = x³ on [0, 2] is:
- 1
- 2/√3
- √3
- 4/3
Show solution
[f(2) − f(0)]/(2 − 0) = 8/2 = 4. So 3c² = 4 ⇒ c = 2/√3 (which lies in (0, 2)).
Answer: (B)
A rectangle has perimeter 40 cm. Find its maximum possible area (in cm²).
Show solution
Let the sides be x and 20 − x. Area A = x(20 − x) = 20x − x². A′ = 20 − 2x = 0 ⇒ x = 10, and A″ = −2 < 0.
Maximum area = 10 × 10 = 100 cm² (a square).
Answer: 100
The volume of the largest right circular cylinder that can be inscribed in a sphere of radius R is:
- 4πR³/(3√3)
- πR³/3
- 2πR³/3
- 4πR³/3
Show solution
Let the cylinder have height h and radius r. From the right triangle through the centre: r² + (h/2)² = R², so r² = R² − h²/4.
V = πr²h = π(R²h − h³/4). V′ = π(R² − 3h²/4) = 0 ⇒ h = 2R/√3, and V″ = −(3π/2)h < 0.
Then r² = R² − R²/3 = 2R²/3, so V = π(2R²/3)(2R/√3) = 4πR³/(3√3).
Answer: (A)
The maximum value of f(x) = 3 sin x + 4 cos x + 2 is:
- 9
- 7
- 5
- 6
Show solution
3 sin x + 4 cos x ranges between −√(3² + 4²) = −5 and 5. So the maximum of f is 5 + 2 = 7.
Answer: (B)
12. Quick Revision Summary
- Tangent slope f′(x1); normal slope −1/f′(x1).
- f′ > 0 increasing, f′ < 0 decreasing; draw a sign chart.
- f′ = 0 and f″ < 0 → max; f″ > 0 → min; check end points for absolute extremes.
- Rolle: f(a) = f(b) ⇒ f′(c) = 0; LMVT: f′(c) = slope of the chord.
- a sin x + b cos x ∈ [−√(a² + b²), √(a² + b²)].