16.1 Limit Concepts and Derivatives of Elementary Functions

16.1 Limit Concepts and Derivatives of Elementary Functions

Differential calculus is one of the most powerful mathematical frameworks tested in the National College Entrance Examination (Gaokao). At its core, calculus transitions mathematical analysis from static quantities to dynamic rates of change. Understanding how limits lead to instantaneous derivatives forms the indispensable foundation for curve analysis, optimization, and real-world mathematical modeling.

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1. Concept of Limits, Continuity, and Average Rate of Change

1.1 Average Rate of Change

Consider a function $y = f(x)$ defined on an open interval containing $x_0$. When the independent variable $x$ changes from $x_0$ to $x_0 + \Delta x$, the corresponding change in the function values is \Delta y = f(x_0 + \Delta x) - f(x_0). The ratio of these increments is defined as the average rate of change of $f(x)$ over the interval $[x_0, x_0 + \Delta x]$:

$\text{Average Rate of Change} = \frac{\Delta y}{\Delta x} = \frac{f(x_0 + \Delta x) - f(x_0)}{\Delta x}$

Geometrically, this ratio represents the secant line slope connecting points $P(x_0, f(x_0))$ and $Q(x_0 + \Delta x, f(x_0 + \Delta x))$ on the graph of $y = f(x)$.

1.2 Instantaneous Rate of Change and the Formal Derivative

As the increment \Delta x shrinks toward $0$, the secant line rotates about point $P$. If the ratio $\frac{\Delta y}{\Delta x}$ approaches a finite limit as \Delta x \to 0, this limit is called the instantaneous rate of change or the derivative of $f(x)$ at $x = x_0$, denoted as $f'(x_0)$ or $\left.\frac{df}{dx}\right|_{x=x_0}$:

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

Alternatively, letting $x = x_0 + \Delta x$, the limit can be expressed as:

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

1.3 Continuity vs. Differentiability

A function $f(x)$ is continuous at $x_0$ if $\lim_{x \to x_0} f(x) = f(x_0)$. A fundamental theorem of calculus establishes that differentiability implies continuity:

$\text{If } f(x) \text{ is differentiable at } x_0, \text{ then } f(x) \text{ is continuous at } x_0.$

Proof: Since $f'(x_0) = \lim_{x \to x_0} \frac{f(x) - f(x_0)}{x - x_0}$, we can write:

$\lim_{x \to x_0} [f(x) - f(x_0)] = \lim_{x \to x_0} \left[ \frac{f(x) - f(x_0)}{x - x_0} \cdot (x - x_0) \right] = f'(x_0) \cdot 0 = 0$

Thus, $\lim_{x \to x_0} f(x) = f(x_0)$, proving continuity. Note that the converse is false: functions like $f(x) = |x|$ are continuous at $x=0$ but non-differentiable because the left-hand limit ($-1$) and right-hand limit ($+1$) of the difference quotient do not match.

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2. Derivation of Standard Elementary Function Derivatives

In Gaokao Mathematics, candidates must memorize and fluently apply the derivative formulas for standard elementary functions. Below are their formal mathematical definitions and key derivations from first principles.

2.1 First-Principles Derivation of Power Function Derivative $f(x) = x^n$ ($n \in \mathbb{N}^*$)

Using the definition of the derivative:

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

Applying the Binomial Expansion Theorem:

$(x + \Delta x)^n = x^n + \binom{n}{1} x^{n-1} \Delta x + \binom{n}{2} x^{n-2} (\Delta x)^2 + \dots + (\Delta x)^n$

Substituting this back into the difference quotient:

$f'(x) = \lim_{\Delta x \to 0} \left[ n x^{n-1} + \binom{n}{2} x^{n-2} \Delta x + \dots + (\Delta x)^{n-1} \right] = n x^{n-1}$

2.2 First-Principles Derivation of Natural Exponential $f(x) = e^x$

Using the standard limit identity $\lim_{t \to 0} \frac{e^t - 1}{t} = 1$:

$f'(x) = \lim_{\Delta x \to 0} \frac{e^{x + \Delta x} - e^x}{\Delta x} = \lim_{\Delta x \to 0} \frac{e^x(e^{\Delta x} - 1)}{\Delta x} = e^x \cdot \lim_{\Delta x \to 0} \frac{e^{\Delta x} - 1}{\Delta x} = e^x \cdot 1 = e^x$

2.3 First-Principles Derivation of Sine Function $f(x) = \sin x$

Using the trigonometric sum-to-product identity $\sin A - \sin B = 2 \cos\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right)$ and important limit $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$:

$f'(x) = \lim_{\Delta x \to 0} \frac{\sin(x + \Delta x) - \sin x}{\Delta x} = \lim_{\Delta x \to 0} \frac{2 \cos\left(x + \frac{\Delta x}{2}\right) \sin\left(\frac{\Delta x}{2}\right)}{\Delta x}$

$= \lim_{\Delta x \to 0} \left[ \cos\left(x + \frac{\Delta x}{2}\right) \cdot \frac{\sin(\Delta x / 2)}{\Delta x / 2} \right] = \cos x \cdot 1 = \cos x$

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3. Differentiation Operational Rules and Composite Chain Rule

Let $u(x)$ and $v(x)$ be differentiable functions on an interval $I$.

1. Linearity (Sum and Constant Multiple Rule): $\left[ c_1 u(x) \pm c_2 v(x) \right]' = c_1 u'(x) \pm c_2 v'(x)$

2. Product Rule: $\left[ u(x) v(x) \right]' = u'(x) v(x) + u(x) v'(x)$

3. Quotient Rule: $\left[ \frac{u(x)}{v(x)} \right]' = \frac{u'(x) v(x) - u(x) v'(x)}{[v(x)]^2} \quad (v(x) \neq 0)$

4. Chain Rule for Composite Functions: If $y = f(u)$ and $u = g(x)$ are differentiable, then the composite function $y = f(g(x))$ has derivative: $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \implies [f(g(x))]' = f'(g(x)) \cdot g'(x)$

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4. Worked Gaokao Exam Problem

Problem Statement

Given the function $f(x) = x^2 e^x - \ln(2x + 1)$ defined on $x > -\frac{1}{2}$:

1. Compute the general derivative expression $f'(x)$.
2. Evaluate the instantaneous rate of change at $x = 0$.
3. Determine the equation of the tangent line to $y = f(x)$ at point $(0, f(0))$.

Step-by-Step Solution

Step 1: Compute $f'(x)$ using differentiation rules. The function consists of two parts: $g(x) = x^2 e^x$ and $h(x) = \ln(2x + 1)$.

• For $g(x) = x^2 e^x$, apply the product rule $(uv)' = u'v + uv'$: $g'(x) = (x^2)' e^x + x^2 (e^x)' = 2x e^x + x^2 e^x = (x^2 + 2x) e^x$
• For $h(x) = \ln(2x + 1)$, apply the chain rule with $u = 2x + 1$ and $h(u) = \ln u$: $h'(x) = \frac{1}{2x + 1} \cdot (2x + 1)' = \frac{2}{2x + 1}$ Combining both components: $f'(x) = (x^2 + 2x)e^x - \frac{2}{2x + 1}$

Step 2: Evaluate $f'(0)$ and $f(0)$.

• Substitute $x = 0$ into $f(x)$: $f(0) = (0)^2 e^0 - \ln(2(0) + 1) = 0 - \ln(1) = 0$
• Substitute $x = 0$ into $f'(x)$: $f'(0) = (0^2 + 2(0))e^0 - \frac{2}{2(0) + 1} = 0 - 2 = -2$

Step 3: Derive the tangent line equation. The point of tangency is $(0, 0)$ and the slope is $k = f'(0) = -2$. Using the point-slope form $y - y_0 = k(x - x_0)$: $y - 0 = -2(x - 0) \implies y = -2x \quad \text{or} \quad 2x + y = 0$

Final Answer: $f'(x) = (x^2 + 2x)e^x - \frac{2}{2x+1}$, $f'(0) = -2$, and the tangent line equation is $2x + y = 0$.