DerivativeCalculus.com

Derivative at a Point Mastery

Master calculating derivatives at specific points with 10 comprehensive problems using limit definition and derivative rules.

Limit Definition Rules Applications

📐 Three Ways to Find f'(a)

1. Limit Definition
f'(a) = limh→0 [f(a+h) - f(a)]/h
or f'(a) = limx→a [f(x) - f(a)]/(x-a)

Use when: Function is simple or you need to demonstrate understanding.

2. Derivative Rules
Find f'(x), then evaluate at x = a

Steps:

  1. Find general derivative f'(x)
  2. Substitute x = a
  3. Simplify

Use when: Function is complex but differentiable.

3. Graphical/Numerical
Slope of tangent line at point

Methods:

  • Estimate from graph
  • Use difference quotient with small h
  • Symmetric difference quotient
Problem 1: Limit Definition (Polynomial) Beginner
f(x) = x² + 3x - 2
Find f'(2) using the limit definition

1 Limit definition:

f'(2) = limh→0 [f(2+h) - f(2)]/h

2 Compute f(2+h):

f(2+h) = (2+h)² + 3(2+h) - 2

= (4 + 4h + h²) + (6 + 3h) - 2

= h² + 7h + 8

3 Compute f(2):

f(2) = 2² + 3(2) - 2 = 4 + 6 - 2 = 8

4 Difference quotient:

[f(2+h) - f(2)]/h = [(h² + 7h + 8) - 8]/h

= (h² + 7h)/h = h + 7

5 Take limit as h→0:

limh→0 (h + 7) = 7

6 Final answer:

f'(2) = 7

Interpretation: At x = 2, the slope of the tangent line is 7. The function is increasing at rate 7 units y per unit x.
💡 Verification: Using derivative rules: f'(x) = 2x + 3, so f'(2) = 2(2) + 3 = 7 ✓
Problem 2: Square Root Function Intermediate
f(x) = √x
Find f'(4) using the limit definition

1 Limit definition:

f'(4) = limx→4 [f(x) - f(4)]/(x-4)

= limx→4 [√x - 2]/(x-4)

2 Rationalize numerator:

Multiply numerator and denominator by conjugate (√x + 2):

= limx→4 [(√x - 2)(√x + 2)]/[(x-4)(√x + 2)]

3 Simplify:

= limx→4 [(x - 4)]/[(x-4)(√x + 2)]

= limx→4 1/(√x + 2)

4 Evaluate limit:

= 1/(√4 + 2) = 1/(2 + 2) = 1/4

5 Final answer:

f'(4) = 1/4 = 0.25

Interpretation: At x = 4, √x increases at rate 0.25. For small changes Δx near 4, √x changes by approximately 0.25·Δx.
💡 General Formula: d/dx[√x] = 1/(2√x), so f'(4) = 1/(2√4) = 1/4 ✓
Problem 3: Trigonometric Function Intermediate
f(x) = sin(x)
Find f'(π/3) using derivative rules

1 Derivative rule for sin(x):

f'(x) = cos(x)

2 Evaluate at x = π/3:

f'(π/3) = cos(π/3)

3 Use known value:

cos(π/3) = 1/2

4 Final answer:

f'(π/3) = 1/2 = 0.5

Geometric Interpretation: At x = π/3 (60°), the slope of sin(x) is 0.5. The tangent line rises 0.5 units vertically for each unit horizontally.
💡 Unit Circle Connection: On unit circle, at angle π/3, coordinates are (1/2, √3/2). The derivative cos(π/3) = x-coordinate = 1/2.
Problem 4: Piecewise Function Advanced
f(x) = { x² if x ≤ 1
2x - 1 if x > 1 }
Determine if f'(1) exists. If so, find it.

1 Check continuity at x = 1:

Left limit: limx→1⁻ f(x) = 1² = 1

Right limit: limx→1⁺ f(x) = 2(1) - 1 = 1

f(1) = 1² = 1

All equal ⇒ f is continuous at x = 1

2 Check differentiability (left derivative):

f'_-(1) = limh→0⁻ [f(1+h) - f(1)]/h

For h<0, f(1+h) = (1+h)² = 1 + 2h + h²

[f(1+h) - f(1)]/h = [(1+2h+h²) - 1]/h = (2h + h²)/h = 2 + h

limh→0⁻ (2 + h) = 2

3 Right derivative:

f'_+(1) = limh→0⁺ [f(1+h) - f(1)]/h

For h>0, f(1+h) = 2(1+h) - 1 = 1 + 2h

[f(1+h) - f(1)]/h = [(1+2h) - 1]/h = 2h/h = 2

limh→0⁺ 2 = 2

4 Compare:

f'_-(1) = 2, f'_+(1) = 2 ⇒ equal

5 Conclusion:

f'(1) exists and f'(1) = 2

Interpretation: Even though the function definition changes at x=1, the slopes from left and right match, so a smooth tangent exists.
💡 Differentiability Test: For piecewise functions at boundary, check: 1. Continuity 2. Left derivative = Right derivative Both must hold for derivative to exist.
Problem 5: Exponential Growth Rate Intermediate
P(t) = 1000·e^(0.05t)
Find the instantaneous growth rate at t = 10 years

1 Derivative represents instantaneous growth rate:

P'(t) = d/dt[1000·e^(0.05t)]

2 Apply chain rule:

P'(t) = 1000·0.05·e^(0.05t) = 50·e^(0.05t)

3 Evaluate at t = 10:

P'(10) = 50·e^(0.05·10) = 50·e^(0.5)

4 Calculate:

e^(0.5) ≈ 1.64872

P'(10) ≈ 50 × 1.64872 ≈ 82.436

5 Interpretation:

At t = 10 years, the population is growing at approximately 82.44 individuals per year.

Relative Growth Rate: The percentage growth rate is constant: P'(t)/P(t) = 0.05 = 5% per year. At t=10: P(10) ≈ 1000·e^(0.5) ≈ 1648.72, and 82.44/1648.72 ≈ 0.05 = 5%.
💡 Exponential Property: For P(t) = P₀·e^(kt), the derivative P'(t) = k·P(t). The growth rate is proportional to the current amount.

📊 Common f'(a) Values to Know

Function f'(x) f'(0) f'(1)
x² 2x 0 2
√x 1/(2√x) ∞ (undefined) 1/2
eˣ eˣ 1 e ≈ 2.718
ln(x) 1/x undefined 1
sin(x) cos(x) 1 cos(1) ≈ 0.540

💪 Additional Practice Problems

Problem 6: f(x) = 1/x at x = 2
Answer: -1/4
Problem 7: f(x) = cos(x) at x = π/2
Answer: -1
Problem 8: f(x) = x³ - 3x at x = -1
Answer: 0 (horizontal tangent)
Problem 9: f(x) = |x| at x = 0 (does derivative exist?)
Answer: No (corner point)

⚠️ Common Mistakes Finding f'(a)