Applications of Derivatives

Why This Chapter Matters

Applications of Derivatives is a high-value JEE topic — 8-12 marks. Maxima/minima, rate of change, tangent/normal equations, and Rolle's/LMVT theorems appear every year.

Core Concepts

1. Rate of Change

If y = f(x), dy/dx = rate of change of y with respect to x.

If x and y both depend on time t: dy/dt = (dy/dx)(dx/dt)

2. Increasing and Decreasing Functions

f is increasing on (a,b) if f'(x) > 0 for all x in (a,b)

f is decreasing on (a,b) if f'(x) < 0 for all x in (a,b)

f'(x) = 0 at critical points (local max/min candidates)

3. Tangent and Normal

For curve y = f(x) at point P(x1, y1):

Slope of tangent = f'(x1)

Equation of tangent: y - y1 = f'(x1)(x - x1)

Equation of normal: y - y1 = -1/f'(x1) x (x - x1)

Tangent parallel to x-axis: f'(x1) = 0

Tangent parallel to y-axis: f'(x1) is undefined

4. Maxima and Minima

First Derivative Test:

At critical point c where f'(c) = 0:

f'(x) changes + to -: local maximum at c

f'(x) changes - to +: local minimum at c

No change: inflection point

Second Derivative Test:

f'(c) = 0 and f''(c) < 0: local maximum

f'(c) = 0 and f''(c) > 0: local minimum

f'(c) = 0 and f''(c) = 0: inconclusive (use first derivative test)

5. Mean Value Theorems

Rolle's Theorem: If f is continuous on [a,b], differentiable on (a,b), and f(a) = f(b),

then there exists c in (a,b) such that f'(c) = 0.

Lagrange's Mean Value Theorem (LMVT):

If f is continuous on [a,b] and differentiable on (a,b),

then there exists c in (a,b) such that f'(c) = [f(b) - f(a)]/(b-a).

(The derivative equals the slope of the secant line at some interior point.)

6. Optimization Problems

Strategy:

1.Define variable to optimize (perimeter, area, volume, cost)

2.Express as function of one variable using constraints

3.Find critical points (set derivative = 0)

4.Check second derivative or endpoints

Classic results:

•Of all rectangles with fixed perimeter P, the square has maximum area: side = P/4

•Of all cylinders with fixed surface area, h = 2r for max volume

•Minimum material for box of given volume: cube

PYQs

2024: Find the point on curve y = x^2 nearest to (0, 5).

Distance^2 = x^2 + (x^2-5)^2. Let u = x^2.

D^2 = u + (u-5)^2 = u^2 - 9u + 25.

d(D^2)/du = 2u - 9 = 0 => u = 9/2 => x^2 = 9/2.

Point: (3/sqrt(2), 9/2).

2023: A particle moves s = t^3 - 3t. Find velocity and acceleration at t=2.

v = ds/dt = 3t^2 - 3 = 3(4)-3 = 9 m/s.

a = dv/dt = 6t = 12 m/s^2.

2022: Find local maxima of f(x) = 2x^3 - 3x^2 - 12x + 4.

f'(x) = 6x^2 - 6x - 12 = 6(x^2-x-2) = 6(x-2)(x+1) = 0 => x = 2 or x = -1.

f''(x) = 12x - 6. f''(-1) = -18 < 0 => local max at x = -1.

f(-1) = -2 - 3 + 12 + 4 = 11.

MCQ Practice

Q1. If f(x) = x^3 - 3x, then f has local minimum at x =

(A) -1 (B) 0 (C) 1 (D) 3

Answer: C [f'(x) = 3x^2-3=0 => x=1 or -1; f''(1)=6>0 => min at x=1]

Q2. Rolle's theorem is applicable to f(x) = |x| on [-1,1]?

No. f is not differentiable at x=0.

Q3 (Hard). A 30m wire is cut into 2 pieces; one bent into square, one into circle. Find lengths to minimize total area.

Let square have side x, so perimeter = 4x, remaining = (30-4x) for circle of circumference 2pir.

r = (30-4x)/(2pi). Area = x^2 + pir^2 = x^2 + (30-4x)^2/(4*pi).

Minimize by differentiating and setting to 0.

Revision Notes

TANGENT at (x1,y1): y - y1 = f'(x1)(x - x1)
NORMAL at (x1,y1): y - y1 = -1/f'(x1) x (x - x1)

CRITICAL POINTS: f'(x) = 0

SECOND DERIVATIVE TEST:
f''(c) < 0 => LOCAL MAX
f''(c) > 0 => LOCAL MIN

ROLLE'S THEOREM: f(a)=f(b) => exists c with f'(c)=0
LMVT: exists c with f'(c) = [f(b)-f(a)]/(b-a)

RATE OF CHANGE:
dy/dt = (dy/dx) x (dx/dt)