导数dǎoshù

Core Concept

The derivative is a core concept in calculus, describing the instantaneous rate of change of a function at a given point. Geometrically, it represents the slope of the tangent line to a curve at that point.

Mathematical Definition

The derivative of function $y = f(x)$ at point $x_0$ is defined as:

$f'(x_0) = \lim_{\Delta x \to 0} \frac{f(x_0 + \Delta x) - f(x_0)}{\Delta x}$

If this limit exists, the function $f(x)$ is said to be differentiable at $x_0$.

Derivative Notation

• $f'(x)$ - Lagrange notation
• $\frac{dy}{dx}$ - Leibniz notation
• $y'$ - abbreviated form
• $\frac{df}{dx}$ - differential form

Common Derivative Formulas

Basic Functions

1. Constant: $(C)' = 0$
2. Power: $(x^n)' = nx^{n-1}$
3. Exponential: $(e^x)' = e^x$, $(a^x)' = a^x \ln a$
4. Logarithmic: $(\ln x)' = \frac{1}{x}$
5. Trigonometric:
   • $(\sin x)' = \cos x$
   • $(\cos x)' = -\sin x$
   • $(\tan x)' = \sec^2 x$

Derivative Rules

1. Sum/Difference: $(f \pm g)' = f' \pm g'$
2. Product: $(fg)' = f'g + fg'$
3. Quotient: $(\frac{f}{g})' = \frac{f'g - fg'}{g^2}$
4. Chain: $(f(g(x)))' = f'(g(x)) \cdot g'(x)$

Applications

1. Finding Tangent Lines

Tangent line to curve $y = f(x)$ at $(x_0, f(x_0))$:

$y - f(x_0) = f'(x_0)(x - x_0)$

2. Determining Monotonicity

• $f'(x) > 0$ → function increasing
• $f'(x) < 0$ → function decreasing
• $f'(x) = 0$ → possible extremum

3. Finding Extrema

Steps:

1. Find derivative $f'(x)$
2. Solve $f'(x) = 0$ for critical points
3. Test sign changes around critical points

CSCA Practice Problems

💡 Note: The following practice problems are designed based on the CSCA exam syllabus and Chinese standardized test formats to help students familiarize themselves with question types and problem-solving approaches.

Example 1: Basic (Difficulty ★★☆☆☆)

Find the derivative of $f(x) = x^3 + 2x$.

Solution: $f'(x) = 3x^2 + 2$

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Example 2: Intermediate (Difficulty ★★★☆☆)

Find the equation of the tangent line to $y = x^2$ at point $(1, 1)$.

Solution:

Step 1: Find derivative $y' = 2x$

Step 2: Find slope at $x=1$: $k = 2(1) = 2$

Step 3: Write tangent equation: $y - 1 = 2(x - 1)$ $y = 2x - 1$

Answer: $y = 2x - 1$

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Example 3: Advanced (Difficulty ★★★★☆)

Find the extrema of $f(x) = x^3 - 3x$.

Solution:

$f'(x) = 3x^2 - 3 = 3(x-1)(x+1)$

Critical points: $x = -1, 1$

• Maximum: $f(-1) = 2$ at $x = -1$
• Minimum: $f(1) = -2$ at $x = 1$

Common Mistakes

❌ Mistake 1: $(x^2)' = 2$

Correction: $(x^2)' = 2x$, not 2! Remember to keep the $x$.

❌ Mistake 2: $(fg)' = f'g'$

Correction: Product rule is $(fg)' = f'g + fg'$, not $f'g'$!

❌ Mistake 3: $f'(x_0) = 0$ always means extremum

Correction: $f'(x_0) = 0$ is only a necessary condition. Must verify sign change.

Study Tips

1. ✅ Understand definition: Derivative = instantaneous rate = tangent slope
2. ✅ Memorize formulas: Learn basic derivatives and rules
3. ✅ Practice: Especially chain rule applications
4. ✅ Applications: Derivatives are widely used in optimization

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💡 Exam Tip: Derivatives account for about 15% of CSCA math questions. Master basic differentiation and geometric applications!