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Geometric Meaning of Derivative

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Geometric Meaning of Derivative

1. Core Concepts

**Geometric Meaning**:

The derivative $f'(x_0)$ of a function $y=f(x)$ at $x_0$ represents the **Slope of the Tangent Line** to the curve at the point $P(x_0, f(x_0))$.

$$ k_{\text{tangent}} = f'(x_0) $$

**Visual Intuition**:

Let $P$ be a fixed point and $Q$ be a moving point on the curve. As $Q$ approaches $P$ infinitely, the **Secant Line** $PQ$ approaches the **Tangent Line**. The derivative quantifies this limit.

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2. Equations of Tangent and Normal Lines

Let the point of tangency be $P(x_0, y_0)$, where $y_0 = f(x_0)$.

#### (1) Tangent Line Equation

Using the point-slope form:

$$ y - y_0 = f'(x_0)(x - x_0) $$

#### (2) Normal Line Equation

The **Normal Line** is perpendicular to the tangent line at the point of tangency.

* **Slope Relationship**: If tangent slope $k = f'(x_0) \neq 0$, then normal slope $k_{\text{norm}} = -\frac{1}{f'(x_0)}$.

* **Equation**:

$$ y - y_0 = -\frac{1}{f'(x_0)}(x - x_0) $$

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3. Typical Examples

**Example 1: Finding Tangent and Normal**

Find the tangent and normal lines to $y = x^2$ at $(1, 1)$.

**Solution**:

1. **Derivative (Slope)**: $y' = 2x$. At $x=1$, tangent slope $k = 2$.

2. **Tangent**: $y - 1 = 2(x - 1) \Rightarrow y = 2x - 1$.

3. **Normal Slope**: $k_{\text{norm}} = -\frac{1}{2}$.

4. **Normal**: $y - 1 = -\frac{1}{2}(x - 1) \Rightarrow y = -\frac{1}{2}x + \frac{3}{2}$.

**Example 2: Finding Tangency Point from Slope**

Find the tangent line to $y = x^3$ with slope 3.

**Solution**:

1. **Set Point**: Let tangency point be $(x_0, y_0)$.

2. **Solve**: $f'(x_0) = 3x_0^2 = 3 \Rightarrow x_0 = \pm 1$.

3. **Two Cases**:

* $x_0 = 1, y_0 = 1 \Rightarrow y = 3x - 2$.

* $x_0 = -1, y_0 = -1 \Rightarrow y = 3x + 2$.

**Example 3: Tangent Passing Through a Point (Trap)**

Find the tangent line to $y = x^2$ passing through $A(2, 0)$.

**Solution**:

*Note: $A(2,0)$ is NOT on the curve. It is an external point.*

1. **Set Tangency Point**: Let point be $P(t, t^2)$.

2. **Slope**: $k = 2t$.

3. **Equation**: $y - t^2 = 2t(x - t)$.

4. **Substitute A**: Plug in $(2, 0)$:

$0 - t^2 = 2t(2 - t) \Rightarrow t^2 - 4t = 0$.

5. **Solve**: $t = 0$ or $t = 4$.

* $t=0 \Rightarrow y=0$.

* $t=4 \Rightarrow y = 8x - 16$.

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4. Common Pitfalls

* **"At the point" vs "Through the point"**:

* "At P": P is the tangency point. Use $f'(x_P)$ directly.

* "Through A": A might not be the tangency point. You MUST **assume a tangency point $(t, f(t))$** first.

* **Normal Slope**: Remember it is the **Negative Reciprocal**.

* **Tangency Condition**: The point satisfies both the curve equation and the line equation.