Application of Derivatives – JEE Main Maths Formula Sheet & Class 12 Notes | All Formulas, Theorems & Results

Tangent and Normal — Equations, Slopes, and Angles Between Curves

Why Tangent and Normal Formulas Are Among the Most Directly Tested Application of Derivatives JEE Main Questions

Tangent to a curve (Application of Derivatives — from PDF): Let y = f(x) be the equation of a curve. At point P(x₁, y₁) on the curve, the value of dy/dx at P gives the slope of the tangent (= tan θ) to the curve at P. This slope is also the slope of the curve at P. If dy/dx does not exist at P (vertical tangent), the tangent is parallel to the y-axis.

Equation of tangent at P(x₁, y₁) to y = f(x) (from PDF — Application of Derivatives):

y – y₁ = (dy/dx)|(x₁,y₁) · (x – x₁)

where (dy/dx)|(x₁,y₁) is the value of dy/dx at the point (x₁, y₁).

Slope of tangent m₁ = f'(x₁). If m₁ = 0 → tangent is parallel to x-axis (horizontal tangent). If m₁ = ∞ (undefined) → tangent is perpendicular to x-axis (vertical tangent, equation: x = x₁).

Equation of normal at P(x₁, y₁) to y = f(x) (from PDF — Application of Derivatives): The normal is perpendicular to the tangent at P. Slope of normal = –1/(dy/dx)|(x₁,y₁) = –1/f'(x₁).

y – y₁ = –1/(dy/dx)|(x₁,y₁) · (x – x₁) [provided dy/dx|(x₁,y₁) ≠ 0]

If dy/dx|(x₁,y₁) = 0 (tangent is horizontal): normal is vertical → equation: x = x₁.

If dy/dx|(x₁,y₁) = ∞ (tangent is vertical): normal is horizontal → equation: y = y₁.

Alternatively from PDF: (y – y₁)·(dy/dx)|(x₁,y₁) + (x – x₁) = 0

Angle between two curves (from PDF — Application of Derivatives JEE Main): Let y = f(x) and y = g(x) intersect at P(x₁, y₁). Slope of tangent Tₓ to y=f(x) at P: m₁ = f'(x₁). Slope of tangent T_g to y=g(x) at P: m₂ = g'(x₁). Angle θ between the curves:

tan θ = |m₁ – m₂| / |1 + m₁m₂| = |f'(x₁) – g'(x₁)| / |1 + f'(x₁)·g'(x₁)|

Remarks from PDF: (1) Curves touch each other (θ = 0) iff m₁ = m₂ iff f'(x₁) = g'(x₁). (2) Curves cut orthogonally (θ = 90°) iff m₁·m₂ = –1 iff f'(x₁)·g'(x₁) = –1. Download the Free PDF for Application of Derivatives for all tangent-normal examples for JEE Main.

Lengths of Tangent, Normal, Sub-Tangent and Sub-Normal — From PDF

Why the Four Length Formulas Are Classic Application of Derivatives JEE Main Direct Questions

Let P(x, y) be any point on the curve y = f(x). The tangent at P meets the x-axis at T, the normal at P meets the x-axis at N, and PS is the perpendicular (ordinate) from P to the x-axis. ST is the sub-tangent and SN is the sub-normal from P. All four length formulas below are directly from the PDF.

Four length formulas (Application of Derivatives — from PDF):

(i) Length of Tangent PT = |y| · √(1 + (dx/dy)²) = |y| · √(1 + 1/(dy/dx)²) = |y/sin θ|

    PT = y · √(1 + (dx/dy)²) where dy/dx is the slope

(ii) Length of Sub-Tangent TS = |y| · |dx/dy| = |y/(dy/dx)|

(iii) Length of Normal PN = |y| · √(1 + (dy/dx)²)

(iv) Length of Sub-Normal SN = |y| · |dy/dx|

Intercepts cut off by tangent (from PDF — Application of Derivatives):

(v) x-intercept of tangent at (x, y) = x – y·(dx/dy) = x – y/(dy/dx)

(vi) y-intercept of tangent at (x, y) = y – x·(dy/dx)

Memory shortcuts for Application of Derivatives lengths (JEE Main): Note that: Sub-tangent TS = y/(dy/dx) (just y over slope). Sub-normal SN = y·(dy/dx) (just y times slope). Normal PN = Sub-tangent × slope² + Sub-tangent = Sub-tangent · sec θ... easier: PN² = y² (1 + m²) where m = dy/dx. These four lengths form two pairs: (Sub-tangent, Sub-normal) and (Tangent, Normal) where product of sub-lengths = y². Download the Free PDF for Application of Derivatives for all length formula examples for JEE Main.

Monotonicity — Increasing and Decreasing Functions

Why Monotonicity Criteria Are the Foundation of All Application of Derivatives JEE Main Questions

Algebraic definition of monotonicity (from PDF — Application of Derivatives): A real-valued function f(x) defined on interval I is: monotonic increasing (non-decreasing) if x₁ < x₂ ⟹ f(x₁) ≤ f(x₂) for all x₁, x₂ ∈ I; strictly increasing if x₁ < x₂ ⟹ f(x₁) < f(x₂). Monotonic decreasing (non-increasing) if x₁ < x₂ ⟹ f(x₁) ≥ f(x₂); strictly decreasing if x₁ < x₂ ⟹ f(x₁) > f(x₂).

Derivative test for monotonicity (from PDF — Application of Derivatives): Let f(x) be continuous on [a, b] and differentiable on (a, b):

(a) f(x) is non-decreasing (increasing) on [a, b] iff f'(x) ≥ 0 for all x ∈ (a, b). It is strictly increasing iff f'(x) > 0 for all x ∈ (a, b).

(b) f(x) is non-increasing (decreasing) on [a, b] iff f'(x) ≤ 0 for all x ∈ (a, b). It is strictly decreasing iff f'(x) < 0 for all x ∈ (a, b).

Important remarks on monotonicity (from PDF — Application of Derivatives JEE Main):

1. A function f(x) is monotonic in [a, b] if it is either entirely increasing or entirely decreasing in [a, b].

2. If f is monotonically increasing on (a, b), then f(x⁺) and f(x⁻) exist at every point x of (a, b). If a < x < y < b then f(x⁺) ≤ f(y⁻) — the right limit at x is ≤ left limit at y.

3. A monotonic function has no discontinuities of the second kind.

4. A strictly increasing or strictly decreasing continuous function is one-one (injective), but the converse is not always true.

5. A monotonic function on [a, b] has at most one zero in [a, b].

Finding intervals of monotonicity (Application of Derivatives — JEE Main standard method): Step 1: Find f'(x). Step 2: Solve f'(x) = 0 to find critical points. Step 3: Determine the sign of f'(x) in each interval between consecutive critical points. Step 4: Where f'(x) > 0 → increasing; f'(x) < 0 → decreasing. This standard method for Application of Derivatives appears in JEE Main as "find intervals where f(x) is increasing/decreasing." Download the Free PDF for Application of Derivatives for all monotonicity examples for JEE Main.

Maxima and Minima — All Three Tests and Critical/Stationary Points

Why Maxima-Minima Is the Highest-Weight Application of Derivatives Section in JEE Main

Definitions (from PDF — Application of Derivatives): x₀ is a point of maximum of f(x) if there exists a neighbourhood U of x₀ such that f(x) ≤ f(x₀) for all x ∈ U (f(x₀) is a local maximum). x₀ is a point of minimum if f(x) ≥ f(x₀) for all x in some neighbourhood of x₀.

Stationary point: x₀ such that f'(x₀) = 0. Critical point: x₀ such that f'(x₀) = 0 OR f'(x₀) does not exist. Maxima and minima can occur only at critical points (necessary condition). Note: not every critical point is a point of extremum.

Necessary condition for extremum (from PDF — Application of Derivatives): x₀ is a point of extremum only if: (i) f'(x₀) = 0, OR (ii) f'(x₀) does not exist but f is continuous at x₀.

Test 1 — First Derivative Test (from PDF — Application of Derivatives JEE Main): Let f(x) be continuous at a critical point x₀:

Case I (Minimum at x₁): If f'(x) changes sign from negative to positive as x crosses x₁ from left to right (f'(x) < 0 for x < x₁ and f'(x) > 0 for x > x₁), then x₁ is a point of local minimum.

Case II (Maximum at x₂): If f'(x) changes sign from positive to negative as x crosses x₂ (f'(x) > 0 for x < x₂ and f'(x) < 0 for x > x₂), then x₂ is a point of local maximum.

Case III (No extremum): If f'(x) does not change sign in moving through the critical point → neither maximum nor minimum (inflection point).

Test 2 — Second Derivative Test (from PDF — Application of Derivatives JEE Main): Let f'(x₀) = 0 and f be twice differentiable at x₀:

If f''(x₀) < 0 → x₀ is a point of local maximum.

If f''(x₀) > 0 → x₀ is a point of local minimum.

If f''(x₀) = 0 → the test is inconclusive (existence of extremum cannot be decided by this condition alone — use higher derivative test or first derivative test).

Test 3 — Higher Order Derivative Test (from PDF — Application of Derivatives): Let n ≥ 2 and f'(x₀) = f''(x₀) = … = f^(n–1)(x₀) = 0, but f^(n)(x₀) ≠ 0:

If n is even: f^(n)(x₀) < 0 → x₀ is a point of local maximum. f^(n)(x₀) > 0 → x₀ is a point of local minimum.

If n is odd: there is no extremum at x₀ (inflection point).

Maxima and Minima for parametric functions (from PDF — Application of Derivatives): Let y = f(x) be given parametrically as x = φ(t), y = ψ(t), both differentiable up to second order, φ'(t) ≠ 0. If at t = t₀, ψ'(t₀) = 0 (dy/dt = 0): (i) If ψ''(t₀) < 0 → local maximum at x = φ(t₀). (ii) If ψ''(t₀) > 0 → local minimum at x = φ(t₀). (iii) If ψ''(t₀) = 0 → inconclusive. Download the Free PDF for Application of Derivatives for all maxima-minima worked examples for JEE Main.

Absolute Maximum and Minimum on a Closed Interval — Application of Derivatives

Why Absolute Extremum on [a,b] Is a Direct JEE Main Application of Derivatives Result

Absolute (global) maximum and minimum (from PDF — Application of Derivatives): The greatest or least value of a continuous function f(x) on [a, b] is attained either at a critical point in (a, b) or at one of the endpoints a or b.

Procedure to find absolute maximum and minimum (from PDF — Application of Derivatives JEE Main):

Step 1: Find all critical points of f in (a, b) — solve f'(x) = 0 and find where f'(x) does not exist.

Step 2: Evaluate f(x) at each critical point and at the two endpoints: f(a) and f(b).

Step 3: Absolute maximum of f on [a, b] = max{f(a), f(b), f at each critical point}.

Absolute minimum of f on [a, b] = min{f(a), f(b), f at each critical point}.

Key note for Application of Derivatives (JEE Main): A function must be continuous on [a, b] for the extreme value theorem (Weierstrass theorem) to guarantee that the absolute maximum and minimum exist. A continuous function on a closed bounded interval is always bounded and always attains its bounds. This is why the domain must be a closed interval [a, b] for absolute extrema, unlike local extrema which are defined in open neighbourhoods.

Distinction between local and absolute extrema (Application of Derivatives — JEE Main): Local (relative) maximum at x₀: f(x₀) ≥ f(x) for x near x₀. Absolute (global) maximum on [a, b]: f(x₀) ≥ f(x) for ALL x ∈ [a, b]. A local maximum value can be less than a local minimum value at a different point. Absolute maximum ≥ all local maxima; absolute minimum ≤ all local minima. Download the Free PDF for Application of Derivatives for absolute extremum examples for JEE Main.

Rolle's Theorem and Lagrange's Mean Value Theorem (LMVT)

Why Rolle's Theorem and LMVT Are Multi-Mark Conceptual Application of Derivatives JEE Main Topics

Rolle's Theorem (from PDF — Application of Derivatives): Let f(x) be a real-valued function defined on [a, b] satisfying three conditions:

(i) f(x) is continuous on [a, b]

(ii) f(x) is differentiable on (a, b) (has a finite derivative at every interior point)

(iii) f(a) = f(b) (equal values at endpoints)

Then there exists at least one c ∈ (a, b) such that f'(c) = 0.

Geometric interpretation of Rolle's Theorem: At the point (c, f(c)) on the graph of y = f(x), the tangent is parallel to the x-axis (horizontal tangent).

Algebraic interpretation of Rolle's Theorem (Application of Derivatives — JEE Main): Between any two zeros of a polynomial, there exists at least one zero of its derivative. If f(a) = f(b), then f'(x) = 0 has at least one real root in (a, b).

Common uses of Rolle's Theorem in JEE Main Application of Derivatives: (1) Proving existence of c where f'(c) = 0. (2) Showing that f(x) = 0 has exactly one root in an interval. (3) Finding the value of the unknown in a polynomial when Rolle's theorem is applicable.

Lagrange's Mean Value Theorem — LMVT (from PDF — Application of Derivatives): Let f(x) be defined on [a, b] such that: (i) f(x) is continuous on [a, b], and (ii) f(x) is differentiable on (a, b). Then there exists at least one c ∈ (a, b) such that:

f'(c) = [f(b) – f(a)] / (b – a)

Geometric interpretation of LMVT: At the point (c, f(c)) on the graph of y = f(x), the tangent is parallel to the chord joining (a, f(a)) and (b, f(b)).

Rearranged: f(b) = f(a) + (b–a)·f'(c) for some c ∈ (a, b). This is also called the finite increment formula.

Test for constancy (from PDF — Application of Derivatives): If f'(x) = 0 for all x in interval I, then f(x) = constant on I. (Consequence of LMVT applied between any two points in I.)

Key connection between Rolle's and LMVT (Application of Derivatives): Rolle's theorem is a special case of LMVT where f(a) = f(b) → [f(b)–f(a)]/(b–a) = 0 → f'(c) = 0. LMVT applies under weaker conditions (no need for f(a) = f(b)). Download the Free PDF for Application of Derivatives for all mean value theorem examples and JEE Main problems.

Rate of Change, Velocity, Acceleration, and Approximations

Why Rate of Change and Approximation Are the Most Applied Application of Derivatives Sections in JEE Main

Rate of Change (from PDF — Application of Derivatives): If y = f(x) is a function of x, and x is a function of time t, then by the chain rule: dy/dt = (dy/dx)·(dx/dt). The ratio dy/dx = (rate of change of y)/(rate of change of x). The derivative dy/dx represents the instantaneous rate of change of y with respect to x.

Velocity and Acceleration (from PDF — Application of Derivatives): If s = distance covered by an object in time t, then:

Velocity v = ds/dt = lim_{Δt→0} (Δs/Δt)

Acceleration a = dv/dt = d²s/dt²

Note: speed = |velocity|. If v > 0 → moving in positive direction; v < 0 → moving in negative direction; v = 0 → object is momentarily at rest (may be a turning point).

Approximate change and approximate value (from PDF — Application of Derivatives JEE Main):

If y = f(x), then the approximate change in y (differential of y) is: Δy ≈ dy = f'(x)·Δx

The approximate value of f at x = a + h (where h is small): f(a + h) ≈ f(a) + h·f'(a)

This is the linearisation formula (first-order Taylor approximation) at x = a.

Working rule for approximation problems (from PDF — Application of Derivatives):

Step 1: Identify a suitable function f(x) matching the given quantity.

Step 2: Find f'(x).

Step 3: Put in f(a + h) = f(a) + h·f'(a). Choose a as the nearest convenient integer (or standard value) and h such that a + h equals the required argument.

Step 4: If finding approximate change in y: use Δy = f'(x)·Δx.

Application of Derivatives rate examples (JEE Main standard problems):

If surface area S = 4πr² and r is increasing at dr/dt = k, then dS/dt = 8πr·(dr/dt) = 8πrk.

If V = (4/3)πr³, then dV/dt = 4πr²·(dr/dt).

If a ladder of length L leans against a wall: x² + y² = L², differentiate w.r.t. t: 2x(dx/dt) + 2y(dy/dt) = 0 → dx/dt = –(y/x)·(dy/dt). These related rates problems are classic Application of Derivatives JEE Main questions. Download the Free PDF for all rate of change and approximation examples.

Mensuration Formulas — All 16 Results for Application of Derivatives Problems

Why Mensuration Formulas Are the Building Blocks of All Application of Derivatives Optimisation Questions in JEE Main

Every Application of Derivatives optimisation problem in JEE Main requires setting up an objective function (what to maximise or minimise) using mensuration formulas. These 16 formulas from the PDF are the direct source material for these problem setups.

All 16 mensuration formulas from PDF — Application of Derivatives:

1. Volume of cuboid = ℓ × b × h (length × breadth × height)

2. Surface area of cuboid = 2(ℓb + bh + hℓ)

3. Volume of cube = a³

4. Surface area of cube = 6a²

5. Volume of cone = (1/3)πr²h (r = base radius, h = height)

6. Curved surface area of cone = πrℓ (ℓ = slant height = √(r²+h²))

7. Curved surface area of cylinder = 2πrh

8. Total surface area of cylinder = 2πrh + 2πr² = 2πr(h + r)

9. Volume of sphere = (4/3)πr³

10. Surface area of sphere = 4πr²

11. Area of circular sector = (1/2)r²θ (θ in radians)

12. Volume of prism = (area of base) × height

13. Lateral surface area of prism = (perimeter of base) × height

14. Total surface area of prism = lateral surface area + 2 × (area of base)

15. Volume of pyramid = (1/3) × (area of base) × height

16. Curved surface area of pyramid = (1/2) × (perimeter of base) × slant height

How mensuration connects to Application of Derivatives optimisation (JEE Main): Example — Cylinder of maximum volume inscribed in a sphere of radius R: Let radius of cylinder = r, height = h. Then r² + (h/2)² = R² (constraint). Volume V = πr²h. Express r² = R² – h²/4. V(h) = π(R² – h²/4)h. Find dV/dh = 0 → h = 2R/√3, r = R√(2/3). Check d²V/dh² < 0 → maximum. These Application of Derivatives optimisation problems using mensuration are direct JEE Main question templates. Download the Free PDF for Application of Derivatives for all optimisation examples with mensuration.