Derivatives Applications – Lesson with Examples

Once you know how to find a derivative, the real power of calculus kicks in — you can use derivatives to analyze functions, solve real-world problems, and understand how things change. This lesson covers every application of derivatives you’ll need for your Calculus 1 final.

📚 Table of Contents

1. Critical Points & Increasing/Decreasing
2. First Derivative Test
3. Concavity & Second Derivative Test
4. Absolute Max & Min on a Closed Interval
5. Optimization Problems
6. Related Rates
7. L’Hôpital’s Rule
8. Tangent Lines & Linear Approximation
9. Common Mistakes to Avoid
10. Practice Problems

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Critical points are where the behavior of a function can change — from increasing to decreasing, or vice versa. They are the starting point for almost every application in this lesson.

How to Find Critical Points

1. Find \( f'(x) \)
2. Set \( f'(x) = 0 \) and solve
3. Also note where \( f'(x) \) is undefined

Increasing & Decreasing

• \( f'(x) > 0 \) on an interval → f is increasing there
• \( f'(x) < 0 \) on an interval → f is decreasing there

✏️ Example 1

Problem: Find the critical points and increasing/decreasing intervals of \( f(x) = x^3 – 6x^2 + 9x \)

Step 1: Find \( f'(x) \):

$$f'(x) = 3x^2 – 12x + 9 = 3(x^2 – 4x + 3) = 3(x-1)(x-3)$$

Step 2: Set \( f'(x) = 0 \): critical points at \( x = 1 \) and \( x = 3 \)

Step 3: Test intervals using a sign chart:

Interval	Test Value	f'(x)	Behavior
\( (-\infty, 1) \)	\( x = 0 \)	+	Increasing ↑
\( (1, 3) \)	\( x = 2 \)	−	Decreasing ↓
\( (3, \infty) \)	\( x = 4 \)	+	Increasing ↑

✅ Answer: Critical points at \( x = 1 \) and \( x = 3 \). Increasing on \( (-\infty, 1) \) and \( (3, \infty) \). Decreasing on \( (1, 3) \).

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Once you have critical points, the First Derivative Test tells you whether each one is a local maximum, local minimum, or neither.

The Rule

• \( f’ \) changes from positive to negative at \( x = c \) → local maximum
• \( f’ \) changes from negative to positive at \( x = c \) → local minimum
• \( f’ \) does not change sign → neither

✏️ Example 2

Problem: Using the result from Example 1, classify the critical points of \( f(x) = x^3 – 6x^2 + 9x \)

From the sign chart above:

• At \( x = 1 \): \( f’ \) goes from + to − → Local Maximum
• At \( x = 3 \): \( f’ \) goes from − to + → Local Minimum

Find the y-values:

$$f(1) = 1 – 6 + 9 = 4 \qquad f(3) = 27 – 54 + 27 = 0$$

✅ Answer: Local max at \( (1, 4) \), local min at \( (3, 0) \)

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The second derivative tells you about the shape of the curve — whether it bends upward (concave up) or downward (concave down).

Concavity Rules

• \( f”(x) > 0 \) → Concave Up (like a bowl ∪)
• \( f”(x) < 0 \) → Concave Down (like a hill ∩)

Inflection Points

An inflection point is where concavity changes. Find them by setting \( f”(x) = 0 \) and checking for a sign change.

Second Derivative Test for Local Extrema

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

• \( f”(c) > 0 \) → Local Minimum
• \( f”(c) < 0 \) → Local Maximum
• \( f”(c) = 0 \) → Inconclusive (use First Derivative Test instead)

✏️ Example 3

Problem: Find concavity and inflection points of \( f(x) = x^4 – 4x^3 \)

Step 1: Find \( f”(x) \):

$$f'(x) = 4x^3 – 12x^2$$ $$f”(x) = 12x^2 – 24x = 12x(x – 2)$$

Step 2: Set \( f”(x) = 0 \): \( x = 0 \) and \( x = 2 \)

Step 3: Sign chart for \( f” \):

Interval	f”(x)	Concavity
\( (-\infty, 0) \)	+	Concave Up ∪
\( (0, 2) \)	−	Concave Down ∩
\( (2, \infty) \)	+	Concave Up ∪

✅ Answer: Inflection points at \( x = 0 \) and \( x = 2 \). Concave up on \( (-\infty, 0) \) and \( (2, \infty) \). Concave down on \( (0, 2) \).

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On a closed interval \( [a, b] \), the absolute maximum and minimum must occur either at a critical point inside the interval or at one of the endpoints. This is called the Closed Interval Method.

Steps

1. Find all critical points of \( f \) on \( (a, b) \)
2. Evaluate \( f \) at each critical point
3. Evaluate \( f \) at both endpoints \( a \) and \( b \)
4. The largest value is the absolute max; the smallest is the absolute min

✏️ Example 4

Problem: Find the absolute max and min of \( f(x) = x^3 – 3x + 2 \) on \( [-2, 3] \)

Step 1: \( f'(x) = 3x^2 – 3 = 3(x-1)(x+1) \)

Critical points: \( x = -1 \) and \( x = 1 \) (both in \( [-2, 3] \))

Step 2: Evaluate at critical points and endpoints:

x	f(x)
\( x = -2 \)	\( -8 + 6 + 2 = 0 \)
\( x = -1 \)	\( -1 + 3 + 2 = 4 \)
\( x = 1 \)	\( 1 – 3 + 2 = 0 \)
\( x = 3 \)	\( 27 – 9 + 2 = 20 \)

✅ Answer: Absolute maximum of 20 at \( x = 3 \). Absolute minimum of 0 at \( x = -2 \) and \( x = 1 \).

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Optimization is about finding the best possible value — maximum area, minimum cost, shortest distance. These are the most common word problems on Calc 1 finals.

5-Step Strategy

1. Draw a diagram and label all variables
2. Write the objective function — what you’re maximizing or minimizing
3. Write the constraint equation — the condition that limits the variables
4. Substitute the constraint into the objective to get one variable
5. Differentiate, set equal to zero, and solve

✏️ Example 5 — Maximum Area

Problem: A farmer has 200 meters of fence to enclose a rectangular field. What dimensions maximize the area?

Constraint: \( 2x + 2y = 200 \Rightarrow y = 100 – x \)

Objective: \( A = xy = x(100 – x) = 100x – x^2 \)

$$A’ = 100 – 2x = 0 \Rightarrow x = 50$$ $$y = 100 – 50 = 50$$

Verify it’s a max: \( A” = -2 < 0 \) ✅

✅ Answer: 50 m × 50 m gives maximum area of 2500 m²

✏️ Example 6 — Minimum Cost

Problem: A box with a square base and open top must hold 32 cubic feet. The base costs $2/ft² and the sides cost $1/ft². Find the dimensions that minimize cost.

Let \( x \) = side of base, \( h \) = height.

Constraint (Volume): \( x^2 h = 32 \Rightarrow h = \frac{32}{x^2} \)

Cost function:

$$C = 2x^2 + 4xh = 2x^2 + 4x \cdot \frac{32}{x^2} = 2x^2 + \frac{128}{x}$$ $$C’ = 4x – \frac{128}{x^2} = 0 \Rightarrow x^3 = 32 \Rightarrow x = \sqrt[3]{32} \approx 3.17 \text{ ft}$$ $$h = \frac{32}{x^2} \approx 3.17 \text{ ft}$$

✅ Answer: Base ≈ 3.17 ft × 3.17 ft, height ≈ 3.17 ft

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Related rates problems involve two or more quantities that are both changing with respect to time. You differentiate an equation with respect to \( t \) and use known rates to find unknown ones.

5-Step Strategy

1. Draw a diagram and label all quantities
2. Identify what’s given and what you need to find
3. Write an equation connecting the variables
4. Differentiate both sides with respect to \( t \)
5. Plug in known values and solve

✏️ Example 7 — Expanding Circle

Problem: A stone is dropped in a pond, creating a circular ripple. The radius is increasing at 3 cm/s. How fast is the area increasing when the radius is 10 cm?

Given: \( \frac{dr}{dt} = 3 \) cm/s. Find: \( \frac{dA}{dt} \) when \( r = 10 \)

Equation: \( A = \pi r^2 \)

Differentiate with respect to t:

$$\frac{dA}{dt} = 2\pi r \frac{dr}{dt} = 2\pi(10)(3) = 60\pi \approx 188.5 \text{ cm}^2/\text{s}$$

✅ Answer: The area is increasing at \( 60\pi \) cm²/s

✏️ Example 8 — Sliding Ladder

Problem: A 10-foot ladder leans against a wall. The bottom slides away at 2 ft/s. How fast is the top sliding down when the bottom is 6 feet from the wall?

Given: \( \frac{dx}{dt} = 2 \) ft/s, ladder = 10 ft. Find: \( \frac{dy}{dt} \) when \( x = 6 \)

Equation: \( x^2 + y^2 = 100 \)

When \( x = 6 \): \( y = \sqrt{100 – 36} = 8 \)

Differentiate with respect to t:

$$2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$$ $$2(6)(2) + 2(8)\frac{dy}{dt} = 0 \Rightarrow \frac{dy}{dt} = -\frac{3}{2} \text{ ft/s}$$

✅ Answer: The top is sliding down at \( \frac{3}{2} \) ft/s

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L’Hôpital’s Rule is a powerful shortcut for limits that produce indeterminate forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \).

The Rule

If \( \lim_{x \to a} \frac{f(x)}{g(x)} \) gives \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), then:

$$\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}$$

⚠️ Differentiate the numerator and denominator separately — this is NOT the Quotient Rule.

✏️ Example 9

Problem: Find \( \lim_{x \to 0} \frac{\sin(x)}{x} \)

Gives \( \frac{0}{0} \). Apply L’Hôpital’s Rule:

$$\lim_{x \to 0} \frac{\cos(x)}{1} = 1$$

✅ Answer: 1

✏️ Example 10

Problem: Find \( \lim_{x \to \infty} \frac{x^2}{e^x} \)

Gives \( \frac{\infty}{\infty} \). Apply twice:

$$\lim_{x \to \infty} \frac{2x}{e^x} \Rightarrow \lim_{x \to \infty} \frac{2}{e^x} = 0$$

✅ Answer: 0

✏️ Example 11

Problem: Find \( \lim_{x \to 0} \frac{e^x – 1}{x} \)

Gives \( \frac{0}{0} \). Apply L’Hôpital’s Rule:

$$\lim_{x \to 0} \frac{e^x}{1} = 1$$

✅ Answer: 1

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The derivative gives the slope of the tangent line at any point. This is one of the most direct and testable applications of derivatives.

Equation of a Tangent Line

At point \( (a, f(a)) \), the tangent line is:

$$y – f(a) = f'(a)(x – a)$$

✏️ Example 12

Problem: Find the tangent line to \( f(x) = x^3 – 2x \) at \( x = 2 \)

Step 1: \( f(2) = 8 – 4 = 4 \) → point \( (2, 4) \)

Step 2: \( f'(x) = 3x^2 – 2 \Rightarrow f'(2) = 10 \)

Step 3:

$$y – 4 = 10(x – 2) \Rightarrow y = 10x – 16$$

✅ Answer: \( y = 10x – 16 \)

Linear Approximation

The tangent line approximates nearby function values:

$$f(x) \approx f(a) + f'(a)(x – a)$$

✏️ Example 13

Problem: Estimate \( \sqrt{9.1} \) using linear approximation

Use \( f(x) = \sqrt{x} \) near \( a = 9 \):

$$f'(x) = \frac{1}{2\sqrt{x}} \Rightarrow f'(9) = \frac{1}{6}$$ $$\sqrt{9.1} \approx 3 + \frac{1}{6}(0.1) \approx 3.0167$$

✅ Answer: ≈ 3.0167

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• ❌ Forgetting to check endpoints in absolute max/min problems on closed intervals
• ❌ Not verifying max vs min — always use the First or Second Derivative Test
• ❌ Related rates: forgetting to differentiate with respect to t — every variable that changes needs a rate
• ❌ Using Quotient Rule for L’Hôpital’s — differentiate top and bottom separately
• ❌ Applying L’Hôpital’s to non-indeterminate forms — check that you have \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) first
• ❌ Optimization: one equation, two unknowns — always use the constraint to eliminate a variable
• ❌ Forgetting units in related rates and optimization answers

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Try each problem before clicking Show Solution!

1. Find the critical points of \( f(x) = 2x^3 – 9x^2 + 12x \)

👁️ Show Solution

$$f'(x) = 6x^2 – 18x + 12 = 6(x-1)(x-2) = 0$$

Critical points at \( x = 1 \) and \( x = 2 \)

2. Find the absolute max and min of \( f(x) = x^3 – 3x^2 \) on \( [-1, 4] \)

👁️ Show Solution

$$f'(x) = 3x^2 – 6x = 3x(x-2) = 0 \Rightarrow x = 0, \; x = 2$$

x	f(x)
−1	−4
0	0
2	−4
4	16

Absolute max: 16 at x = 4. Absolute min: −4 at x = −1 and x = 2.

3. Find the inflection points of \( f(x) = x^3 – 6x^2 + 9x + 1 \)

👁️ Show Solution

$$f”(x) = 6x – 12 = 0 \Rightarrow x = 2$$

\( f(2) = 8 – 24 + 18 + 1 = 3 \)

Inflection point at \( (2, 3) \)

4. Find the tangent line to \( f(x) = x^2 + 3x \) at \( x = 1 \)

👁️ Show Solution

\( f(1) = 4 \) → point \( (1, 4) \)

\( f'(x) = 2x + 3 \Rightarrow f'(1) = 5 \)

$$y – 4 = 5(x-1) \Rightarrow y = 5x – 1$$

5. Find \( \lim_{x \to 0} \frac{\tan(x)}{x} \) using L’Hôpital’s Rule

👁️ Show Solution

$$\lim_{x \to 0} \frac{\sec^2(x)}{1} = 1$$

6. A rectangle has a perimeter of 80 cm. Find the dimensions that maximize the area.

👁️ Show Solution

\( y = 40 – x \), \( A = x(40-x) \)

$$A’ = 40 – 2x = 0 \Rightarrow x = 20, \; y = 20$$

20 cm × 20 cm

7. A spherical balloon is inflated so the radius increases at 2 cm/s. How fast is the volume increasing when \( r = 5 \) cm?

👁️ Show Solution

$$\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} = 4\pi(25)(2) = 200\pi \approx 628.3 \text{ cm}^3/\text{s}$$

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