Applications of Derivatives — Class 12

Introduction

English: This chapter focuses on how derivatives are used to solve practical problems: finding tangents and normals, locating maxima and minima, optimization in geometry and economics, analysing increasing/decreasing intervals, concavity and points of inflection, and sketching curves. Intuitive diagrams, theorem proofs, solved examples and a strong practice set are included. SEO keywords embedded: Class 12 Maths, applications of derivatives, maxima minima, optimization, tangent, normal, rsetu.link.

Marathi / मराठी (संक्षेप): हा अध्याय derivatives चा उपयोग शिकवतो: tangent/normal काढणे, maxima/minima, optimization problems, curve sketching, इत्यादी. सर्व सिद्धांत व उदाहरणे सोप्या टप्प्यात देण्यात आहे.

Topics Covered

• Tangent and normal to a curve
• Gradient and geometric meaning of derivative
• Increasing / decreasing functions
• Local and absolute maxima / minima
• First and second derivative tests
• Concavity and point of inflection
• Optimization problems (geometry, economics)
• Curve sketching steps and examples
• Approximation & linearization

Key Definitions

1. Tangent at (a,f(a)): Line with slope f'(a), equation: y - f(a) = f'(a)(x-a).
2. Normal: Perpendicular to tangent; slope = -1/f'(a) (if f'(a)≠0).
3. Critical point: x where f'(x)=0 or f' undefined; candidates for local extrema.
4. Inflection point: Point where concavity changes; f'' changes sign.

Important Theorems (with proofs)

1. First derivative test

If f'(x) changes sign from + to − at c, f has local max at c; from − to + gives local min. Proof based on monotonicity on intervals.

2. Second derivative test

If f'(c)=0 and f''(c)>0 then local minimum; if f''(c)<0 then local maximum. Proof via Taylor expansion / quadratic approximation.

3. Test for point of inflection

If f'' changes sign at c and f is continuous, c is point of inflection. Note: f''(c)=0 is necessary but not sufficient.

Methods & Shortcuts

• To sketch: find domain → intercepts → critical points → monotonicity → concavity → asymptotes → plot.
• Optimization: express quantity to minimize/maximize in single variable, differentiate, solve f'=0, test endpoints.
• Linear approximation: f(a+h) ≈ f(a) + f'(a)h for small h.

Solved Examples (8) — step-by-step

1. Example 1: Find tangent and normal to y=x^2 at x=1.

   f'(x)=2x ⇒ slope at 1 is 2. Tangent: y-1=2(x-1) ⇒ y=2x-1. Normal slope = -1/2 ⇒ y-1=-1/2(x-1).

2. Example 2: Maximize area of rectangle inscribed under parabola y=4-x^2.

   Width 2x, height y=4-x^2 ⇒ area A(x)=2x(4-x^2)=8x-2x^3. A'(x)=8-6x^2=0 ⇒ x=±√(4/3). Take positive. A''(x)=-12x ⇒ negative ⇒ local max.

3. Example 3: Find intervals where f(x)=x^3-3x^2+2 is increasing.

   f'(x)=3x^2-6x=3x(x-2). Roots at 0 and 2. Test sign: (-∞,0):+, (0,2):-, (2,∞):+ ⇒ increasing on (-∞,0) and (2,∞).

4. Example 4: Show that curve y= x^3 has inflection at 0.

   f''(x)=6x; changes sign at 0 ⇒ inflection point at origin.

5. Example 5 (Optimization): Minimum cost: Given perimeter fixed, find rectangle of max area? (Classic derivation)

   For fixed perimeter P=2(l+w), maximize A=lw ⇒ w=(P/2)-l ⇒ A(l)=l(P/2 - l) ⇒ differentiate → l=P/4; square is optimal.

6. Example 6 (Economics): Find production level maximizing profit π(q)=R(q)-C(q) where R=q(100-0.5q), C=20q+200.

   π(q)=100q-0.5q^2-20q-200=80q-0.5q^2-200. π'(q)=80-q ⇒ set 0 ⇒ q=80. π''(q)=-1 <0 ⇒ max.

7. Example 7 (Approximation): Approximate √(4.04) using linearization around 4.

   f(x)=√x, f'(x)=1/(2√x). At a=4, f(a)=2, f'(a)=1/4 ⇒ Δx=0.04 ⇒ f≈2 + (1/4)(0.04)=2.01.

8. Example 8 (Curve sketching): Sketch y=(x+1)/(x^2+1) — find critical points, asymptotes and behaviour.

   Compute f', f'', find extrema and inflection, note horizontal asymptote y~0 as x→±∞, intercepts at (-1,0) etc. (Detailed steps included in full notes).

Practice Questions (12)

1. Find tangent to y=sin x at x=π/6.
2. Maximize volume of closed box with square base and given surface area.
3. Find local extrema of f(x)=x^4-4x^3+6x^2.
4. Where is f(x)=x/(1+x^2) increasing?
5. Show that the curve y=x^4 has an inflection point at x=0?
6. Optimize fence problem: Given 100m fencing, find rectangle dimensions for max area.
7. Find normal line to y=ln x at x=1.
8. Using linearization estimate (1.02)^5.
9. Find point(s) where tangent is horizontal for y= x^3 - 3x +1.
10. Show that for f(x)=e^{-x^2}, f has maximum at x=0.
11. Sketch y=x^3-3x and identify local extrema and inflection.
12. Use second derivative test to classify critical points of f(x)=x^3-6x^2+9x.

Summary Table (Quick Revision)

Answer Key — Practice Qs (brief)

1. Tangent slope cos(π/6)=√3/2 ⇒ equation y-1/2=(√3/2)(x-π/6).
2. Square base with side = √(surface area/6) etc. (worked steps in full notes).
3. Compute f', solve roots and test.
4. Compute derivative and sign analysis.
5. f''(x)=12x^2; at 0 f''=0 but f''' shows change in concavity — discuss caution. (Detailed justification in notes.)
6. 50m by 25m (for 100m perimeter rectangle maximizing area is 25x25? Wait compute: for fixed perimeter P=100, square with side 25.)
7. Normal at x=1 for ln x: slope of tangent =1 ⇒ normal slope = -1 ⇒ eqn: y-0 = -1(x-1).
8. Use binomial expansion or linearization: (1.02)^5 ≈ 1 + 5(0.02)=1.10 (first-order approx).
9. Set derivative 3x^2 -3 =0 ⇒ x=±1 ⇒ check horizontal tangents.
10. f'(x) = -2x e^{-x^2}; zero at 0 and f'' negative at 0 → maximum at 0.
11. Find derivative and f'', classify.
12. Compute f' and f'' and apply second derivative test.