Application of Derivatives

Rate of Change of Quantities

**Definition:** When a quantity y varies with respect to another quantity x satisfying y = f(x), then dy/dx (or f'(x)) represents the **rate of change of y with respect to x**. At a specific point x = x₀, the rate of change is dy/dx|ₓ₌ₓ₀ = f'(x₀).

**Key Principle:** If two variables x and y both depend on a third variable t (i.e., x = f(t) and y = g(t)), then by the Chain Rule:

dy/dx = (dy/dt)/(dx/dt), provided dx/dt ≠ 0

This allows calculation of rates of change between two variables using their individual rates of change with respect to a common parameter.

**Important Note:** dy/dx is positive when y increases as x increases, and negative when y decreases as x increases.

**Example 1: Area of Circle**

Find the rate of change of area of a circle with respect to radius r when r = 5 cm.

Given: A = πr²

dA/dr = 2πr

At r = 5 cm: dA/dr = 2π(5) = 10π cm²/cm

**Example 2: Cube Volume and Surface Area**

Volume of cube increases at 9 cm³/s. Find how fast surface area increases when edge = 10 cm.

Given: V = x³, S = 6x²

dV/dt = 9

From V: dV/dt = 3x²(dx/dt), so 9 = 3x²(dx/dt)

Therefore: dx/dt = 3/x²

Now: dS/dt = 12x(dx/dt) = 12x · (3/x²) = 36/x

At x = 10: dS/dt = 36/10 = 3.6 cm²/s

**Example 3: Related Rates with Circles**

Stone creates circular waves at 4 cm/s. When radius = 10 cm, how fast is enclosed area increasing?

Given: A = πr², dr/dt = 4 cm/s

dA/dt = 2πr(dr/dt) = 2π(10)(4) = 80π cm²/s

**Example 4: Rectangle Dimensions Changing**

Length x decreases at 3 cm/min, width y increases at 2 cm/min. When x = 10 cm, y = 6 cm, find rates of change of perimeter and area.

Given: dx/dt = -3 cm/min, dy/dt = 2 cm/min

(a) Perimeter: P = 2(x + y)

dP/dt = 2(dx/dt + dy/dt) = 2(-3 + 2) = -2 cm/min

(b) Area: A = xy

dA/dt = y(dx/dt) + x(dy/dt) = 6(-3) + 10(2) = -18 + 20 = 2 cm²/min

**Economic Applications:**

**Marginal Cost:** MC = dC/dx represents the instantaneous rate of change of total cost with respect to output level.

**Marginal Revenue:** MR = dR/dx represents the rate of change of total revenue with respect to number of items sold.

**Example 5: Marginal Cost**

C(x) = 0.005x³ - 0.02x² + 30x + 5000

dC/dx = 0.015x² - 0.04x + 30

At x = 3: MC = 0.015(9) - 0.04(3) + 30 = 0.135 - 0.12 + 30 = 30.015 rupees

Increasing and Decreasing Functions

**Definition 1 (Monotonicity on Interval):** Let I be an interval in the domain of f.

**Increasing on I:** x₁ < x₂ in I ⟹ f(x₁) < f(x₂) for all x₁, x₂ ∈ I

**Decreasing on I:** x₁ < x₂ in I ⟹ f(x₁) > f(x₂) for all x₁, x₂ ∈ I

**Constant on I:** f(x) = c for all x ∈ I

**Strictly Increasing:** Same as increasing with strict inequality

**Strictly Decreasing:** Same as decreasing with strict inequality

**Definition 2 (Monotonicity at a Point):** f is increasing (or decreasing) at x₀ if there exists an open interval containing x₀ where f is increasing (or decreasing).

**Theorem 1 (First Derivative Test for Monotonicity):** Let f be continuous on [a,b] and differentiable on (a,b). Then:

(a) f is increasing in [a,b] if f'(x) > 0 for each x ∈ (a,b)

(b) f is decreasing in [a,b] if f'(x) < 0 for each x ∈ (a,b)

(c) f is constant in [a,b] if f'(x) = 0 for each x ∈ (a,b)

**Proof of (a):** By Mean Value Theorem, for x₁ < x₂ in [a,b], there exists c ∈ (x₁, x₂) such that f(x₂) - f(x₁) = f'(c)(x₂ - x₁). Since f'(c) > 0 and (x₂ - x₁) > 0, we get f(x₂) > f(x₁).

**Procedure to Find Intervals of Increase/Decrease:**

1. Find f'(x)

2. Set f'(x) = 0 and solve for critical points

3. Partition the domain using critical points

4. Check sign of f'(x) in each interval

5. If f'(x) > 0, function is increasing; if f'(x) < 0, function is decreasing

**Example 6: Polynomial Function**

f(x) = x² - 4x + 6

f'(x) = 2x - 4 = 0 ⟹ x = 2

For x < 2: f'(x) = 2(x-2) < 0, so f is decreasing on (-∞, 2)

For x > 2: f'(x) > 0, so f is increasing on (2, ∞)

**Example 7: Higher Degree Polynomial**

f(x) = 4x³ - 6x² - 72x + 30

f'(x) = 12x² - 12x - 72 = 12(x² - x - 6) = 12(x - 3)(x + 2)

Critical points: x = -2, 3

| Interval | Sign of f'(x) | Nature |

|----------|---------------|---------|

| (-∞, -2) | (+)(−) = (−) < 0 | Decreasing |

| (-2, 3) | (−)(−) = (+) > 0 | Increasing |

| (3, ∞) | (+)(+) = (+) > 0 | Increasing |

Wait, recalculating: at x = -3: f'(-3) = 12(-6)(-1) = 72 > 0, at x = 0: f'(0) = 12(-3)(2) = -72 < 0, at x = 4: f'(4) = 12(1)(6) = 72 > 0

Correct: f increases on (-∞, -2), decreases on (-2, 3), increases on (3, ∞)

**Example 8: Trigonometric Function**

f(x) = sin 3x, x ∈ [0, 2π]

f'(x) = 3cos 3x = 0 ⟹ 3x = π/2, 3π/2 ⟹ x = π/6, π/2

For x ∈ [0, π/6]: cos 3x > 0, so f'(x) > 0, function increasing

For x ∈ [π/6, π/2]: cos 3x < 0, so f'(x) < 0, function decreasing

**Example 9: Combination Function**

f(x) = sin x + cos x, 0 ≤ x ≤ 2π

f'(x) = cos x - sin x = 0 ⟹ tan x = 1 ⟹ x = π/4, 5π/4

[0, π/4]: cos x > sin x, so f'(x) > 0, increasing

[π/4, 5π/4]: cos x < sin x, so f'(x) < 0, decreasing

[5π/4, 2π]: cos x > sin x, so f'(x) > 0, increasing

Maxima and Minima

**Definitions:**

**Local Maximum:** f has local maximum at x = c if f(c) ≥ f(x) for all x in some neighborhood of c

**Local Minimum:** f has local minimum at x = c if f(c) ≤ f(x) for all x in some neighborhood of c

**Global (Absolute) Maximum:** f has global maximum at x = c if f(c) ≥ f(x) for all x in domain

**Global (Absolute) Minimum:** f has global minimum at x = c if f(c) ≤ f(x) for all x in domain

**Turning Points (Critical Points):** Points where f'(x) = 0 or f'(x) is undefined

**Theorem 2 (Fermat's Theorem):** If f has a local maximum or minimum at c and f'(c) exists, then f'(c) = 0.

Note: Converse is not always true. f'(c) = 0 does not guarantee local extremum.

**First Derivative Test for Local Extrema:**

If c is a critical point and f is continuous at c:

If f'(x) changes from positive to negative at c, then f has a **local maximum** at c

If f'(x) changes from negative to positive at c, then f has a **local minimum** at c

If f'(x) does not change sign, then c is **neither maximum nor minimum** (inflection point)

**Procedure:**

1. Find all critical points by solving f'(x) = 0

2. Check sign of f'(x) on either side of each critical point

3. Apply first derivative test

**Example 10: First Derivative Test**

f(x) = 4x³ - 6x² - 72x + 30

f'(x) = 12(x + 2)(x - 3)

Critical points: x = -2, 3

| Interval | Sign of f'(x) | Behavior |

|----------|---------------|----------|

| x < -2 | (+) | Increasing |

| -2 < x < 3 | (−) | Decreasing |

| x > 3 | (+) | Increasing |

At x = -2: f'(x) changes from (+) to (−), so **local maximum**

At x = 3: f'(x) changes from (−) to (+), so **local minimum**

**Second Derivative Test:**

If f is continuous at critical point c where f'(c) = 0:

If f''(c) < 0, then f has a **local maximum** at c

If f''(c) > 0, then f has a **local minimum** at c

If f''(c) = 0, test is **inconclusive**

**Example 11: Second Derivative Test**

f(x) = x³ - 3x² + 3x - 100

f'(x) = 3x² - 6x + 3 = 3(x - 1)² = 0 ⟹ x = 1

f''(x) = 6x - 6 = 6(x - 1)

f''(1) = 0 ⟹ test is inconclusive

Checking first derivative: f'(x) = 3(x-1)² ≥ 0 for all x, and equals 0 only at x = 1. Function is increasing, so x = 1 is neither maximum nor minimum.

**Finding Absolute Maximum and Minimum on Closed Interval [a,b]:**

1. Find all critical points c in (a,b) where f'(c) = 0

2. Evaluate f at all critical points and endpoints

3. The largest value is absolute maximum, smallest is absolute minimum

**Example 12: Absolute Extrema**

f(x) = 3x⁴ - 4x³ - 12x² + 5 on [-2, 2]

f'(x) = 12x³ - 12x² - 24x = 12x(x² - x - 2) = 12x(x - 2)(x + 1)

Critical points in (-2, 2): x = -1, 0

Evaluate:

f(-2) = 3(16) - 4(-8) - 12(4) + 5 = 48 + 32 - 48 + 5 = 37

f(-1) = 3(1) - 4(-1) - 12(1) + 5 = 3 + 4 - 12 + 5 = 0

f(0) = 5

f(2) = 3(16) - 4(8) - 12(4) + 5 = 48 - 32 - 48 + 5 = -27

**Absolute maximum:** f(-2) = 37

**Absolute minimum:** f(2) = -27

**Important Distinctions:**

**Local extremum** refers to comparison with nearby points

**Absolute extremum** refers to entire domain

**Critical point** where f'(x) = 0 or undefined, but may not be extremum

**Stationary point** where f'(x) = 0 exactly

**Inflection point** where concavity changes but may not have f'(x) = 0

Tangent and Normal to a Curve

**Tangent Line:** The tangent to curve y = f(x) at point (x₀, y₀) is a straight line that touches the curve at exactly that point.

**Slope of Tangent:** At point (x₀, y₀) on curve y = f(x), slope of tangent = f'(x₀) = dy/dx|ₓ₌ₓ₀

**Equation of Tangent:** At point (x₀, y₀):

**y - y₀ = f'(x₀)(x - x₀)**

Or: **y - y₀ = m(x - x₀)** where m = f'(x₀)

**Normal Line:** The line perpendicular to the tangent at the point of contact.

**Slope of Normal:** If slope of tangent = m, then slope of normal = -1/m (for m ≠ 0)

**Equation of Normal:** At point (x₀, y₀):

**y - y₀ = -1/f'(x₀) (x - x₀)**

Or: **y - y₀ = -1/m (x - x₀)**

**Special Cases:**

If f'(x₀) = 0, tangent is horizontal (y = y₀) and normal is vertical (x = x₀)

If f'(x₀) is undefined (vertical tangent), tangent is vertical (x = x₀) and normal is horizontal (y = y₀)

**Example 13: Tangent and Normal**

Find equations of tangent and normal to y = x² - 3x + 2 at point x = 1.

At x = 1: y = 1 - 3 + 2 = 0, so point is (1, 0)

f'(x) = 2x - 3

f'(1) = 2 - 3 = -1

Slope of tangent = -1

**Tangent:** y - 0 = -1(x - 1) ⟹ **y = -x + 1** or **x + y - 1 = 0**

Slope of normal = -1/(-1) = 1

**Normal:** y - 0 = 1(x - 1) ⟹ **y = x - 1** or **x - y - 1 = 0**

**Example 14: Horizontal Tangent**

Find point on curve y = x³ - 3x where tangent is horizontal.

f'(x) = 3x² - 3 = 0 ⟹ x² = 1 ⟹ x = ±1

At x = 1: y = 1 - 3 = -2, point is (1, -2), tangent is **y = -2**

At x = -1: y = -1 + 3 = 2, point is (-1, 2), tangent is **y = 2**

Approximations Using Differentials

**Differential:** For small change Δx in x, the corresponding change in y is Δy = f(x + Δx) - f(x).

The differential dy ≈ f'(x)·Δx provides an approximation of Δy.

**Linear Approximation:**

**f(x + Δx) ≈ f(x) + f'(x)·Δx**

Or: **f(a + h) ≈ f(a) + f'(a)·h** for small h

**Key Principle:** For small Δx, the change in function value can be approximated by the derivative times the change in variable.

**Example 15: Approximation using Differential**

Approximate √26 using differential.

Let f(x) = √x, x = 25, Δx = 1

f(x) = √x, f'(x) = 1/(2√x)

f(25) = 5, f'(25) = 1/10

√26 = f(25 + 1) ≈ f(25) + f'(25)·(1)

√26 ≈ 5 + (1/10)(1) = 5 + 0.1 = **5.1**

Actual value ≈ 5.0990..., so approximation is very accurate.

**Example 16: Approximation of Cube Root**

Find approximate value of ∛65.

Let f(x) = ∛x = x^(1/3), a = 64, h = 1

f(64) = 4, f'(x) = (1/3)x^(-2/3)

f'(64) = (1/3)(64)^(-2/3) = (1/3)(1/16) = 1/48

∛65 ≈ 4 + (1/48)(1) = 4 + 1/48 ≈ **4.021**

Optimization Problems (Real-life Applications)

**Strategy for Solving Optimization Problems:**

1. **Identify the quantity** to be maximized or minimized

2. **Express it as function** of one variable (eliminate other variables using constraints)

3. **Find domain** of the function

4. **Find critical points** by setting derivative = 0

5. **Check endpoints** if domain is closed

6. **Evaluate function** at critical points and endpoints

7. **Interpret results** in context of problem

**Example 17: Manufacturing Problem**

A manufacturer wants to design a cylindrical tin can with volume 1000 cm³ to minimize surface area (and thus cost).

Volume: V = πr²h = 1000 ⟹ h = 1000/(πr²)

Surface area: S = 2πr² + 2πrh = 2πr² + 2πr · 1000/(πr²) = 2πr² + 2000/r

To minimize: dS/dr = 4πr - 2000/r² = 0

4πr = 2000/r²

4πr³ = 2000

r³ = 500/π

r = (500/π)^(1/3) cm

Check second derivative: d²S/dr² = 4π + 4000/r³ > 0 for all r > 0, confirming minimum.

h = 1000/(πr²) = 2r (at optimal point)

**Optimal dimensions:** height = 2 × radius

**Example 18: Profit Maximization**

Profit function: P(x) = -2x² + 100x - 500, where x is number of units

dP/dx = -4x + 100 = 0 ⟹ x = 25 units

d²P/dx² = -4 < 0, confirming maximum

Maximum profit = P(25) = -2(625) + 100(25) - 500 = -1250 + 2500 - 500 = **750 rupees**

**Example 19: Ladder Problem**

A 5m ladder leans against a wall. Bottom pulled at 2 cm/s away from wall. How fast is height decreasing when foot is 4m from wall?

Constraint: x² + y² = 25 (where x is distance from wall, y is height)

Differentiating: 2x(dx/dt) + 2y(dy/dt) = 0

When x = 4: y = √(25-16) = 3m

2(4)(2) + 2(3)(dy/dt) = 0

16 + 6(dy/dt) = 0

dy/dt = -8/3 cm/s

Height decreasing at **8/3 cm/s** ≈ 2.67 cm/s

Concavity and Inflection Points

**Concavity:** A function f is:

**Concave up (convex)** on interval I if f''(x) > 0 for all x ∈ I (graph curves upward)

**Concave down (concave)** on interval I if f''(x) < 0 for all x ∈ I (graph curves downward)

**Inflection Point:** A point x = c where f''(x) changes sign is an inflection point. The graph changes concavity at this point.

**Procedure to Find Inflection Points:**

1. Find f''(x)

2. Set f''(x) = 0 and solve for critical points

3. Check if f''(x) changes sign around each point

4. If sign changes, it's an inflection point

**Example 20: Inflection Points**

f(x) = x³ - 3x² + 3x - 5

f'(x) = 3x² - 6x + 3

f''(x) = 6x - 6 = 0 ⟹ x = 1

For x < 1: f''(x) < 0, concave down

For x > 1: f''(x) > 0, concave up

At x = 1: f''(x) changes from negative to positive, so **x = 1 is inflection point**

y-coordinate: f(1) = 1 - 3 + 3 - 5 = -4

**Inflection point:** (1, -4)

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**KEY EXAMINATION POINTS:**

Rate of change problems require proper use of chain rule and identification of related variables

Always find critical points correctly when analyzing monotonicity

First derivative test requires checking sign change, not just the value of f'(x)

Second derivative test is faster but inconclusive when f''(c) = 0

In optimization, always consider boundary points and domain restrictions

Tangent and normal equations must be at specific given points

Approximation using differentials works best for small changes

Real-life problems require careful variable identification and constraint formulation