Derivative as Slope

Understanding derivatives through tangent lines and slope

The Derivative as Slope

The most visual way to understand derivatives: they give you the slope of the tangent line!

Tangent Lines Explained

A tangent line to a curve at a point:

  • Touches the curve at exactly that point
  • Has the same "direction" as the curve at that point
  • Doesn't cross the curve (locally)

Think of it as the "best linear approximation" to the curve at that point.

The Connection

slope of tangent line at x=a=f(a)\text{slope of tangent line at } x = a \quad = \quad f'(a)

The derivative is the slope!

Finding the Tangent Line Equation

Once you know the slope m=f(a)m = f'(a) and the point (a,f(a))(a, f(a)), use point-slope form:

yf(a)=f(a)(xa)y - f(a) = f'(a)(x - a)

Or rearrange to slope-intercept form:

y=f(a)(xa)+f(a)y = f'(a)(x - a) + f(a)

Example 1: Find the Tangent Line

Find the equation of the tangent line to f(x)=x2f(x) = x^2 at x=2x = 2.

Step 1: Find the point f(2)=22=4f(2) = 2^2 = 4 Point: (2,4)(2, 4)

Step 2: Find the slope (derivative) Using the power rule (we'll learn this soon): f(x)=2xf'(x) = 2x f(2)=2(2)=4f'(2) = 2(2) = 4 Slope: m=4m = 4

Step 3: Use point-slope form y4=4(x2)y - 4 = 4(x - 2) y4=4x8y - 4 = 4x - 8 y=4x4y = 4x - 4

Answer: The tangent line is y=4x4y = 4x - 4

Positive, Negative, and Zero Slopes

The derivative tells you the curve's behavior:

| f(a)f'(a) | Tangent Line | Curve Behavior | |---------|--------------|----------------| | f(a)>0f'(a) > 0 | Slopes upward | Increasing | | f(a)<0f'(a) < 0 | Slopes downward | Decreasing | | f(a)=0f'(a) = 0 | Horizontal | Flat (critical point) |

Critical Points

When f(a)=0f'(a) = 0, we have a critical point:

  • Could be a maximum (top of a hill)
  • Could be a minimum (bottom of a valley)
  • Could be a saddle point (neither)

The horizontal tangent line is a clue that something interesting is happening!

Example 2: Horizontal Tangent

Find where f(x)=x24x+3f(x) = x^2 - 4x + 3 has a horizontal tangent line.

Step 1: Find the derivative f(x)=2x4f'(x) = 2x - 4

Step 2: Set equal to zero 2x4=02x - 4 = 0 2x=42x = 4 x=2x = 2

Step 3: Find the point f(2)=48+3=1f(2) = 4 - 8 + 3 = -1

Answer: Horizontal tangent at (2,1)(2, -1)

This is actually the vertex (minimum) of the parabola!

Secant Lines vs. Tangent Lines

Secant line: Connects two points on the curve

  • Slope: f(b)f(a)ba\frac{f(b) - f(a)}{b - a} (average rate of change)

Tangent line: Touches at one point

  • Slope: f(a)f'(a) (instantaneous rate of change)

As the two points of a secant line get closer, the secant line approaches the tangent line!

Normal Lines

The normal line is perpendicular to the tangent line.

If the tangent line has slope m=f(a)m = f'(a), the normal line has slope:

mnormal=1f(a)m_{\text{normal}} = -\frac{1}{f'(a)}

(Negative reciprocal)

Graphical Analysis

Looking at a graph:

  • Steep tangent → Large f(a)|f'(a)| → Fast change
  • Gentle tangent → Small f(a)|f'(a)| → Slow change
  • Horizontal tangentf(a)=0f'(a) = 0 → No change (momentarily)

Practice Visualization

For any curve:

  1. Pick a point
  2. Imagine a line that "just kisses" the curve there
  3. That line's slope is the derivative
  4. If you can't draw it without lifting your pencil, it's not differentiable there!

📚 Practice Problems

1Problem 1medium

Question:

Find the equation of the tangent line to f(x)=x3f(x) = x^3 at the point (1,1)(1, 1).

💡 Show Solution

Step 1: Verify the point f(1)=13=1f(1) = 1^3 = 1 ✓ The point (1,1)(1, 1) is on the curve.

Step 2: Find the derivative Using the power rule: f(x)=3x2f'(x) = 3x^2

Step 3: Find the slope at x = 1 f(1)=3(1)2=3f'(1) = 3(1)^2 = 3

Step 4: Use point-slope form y1=3(x1)y - 1 = 3(x - 1) y1=3x3y - 1 = 3x - 3 y=3x2y = 3x - 2

Answer: y=3x2y = 3x - 2

We can verify: at x=1x = 1, y=3(1)2=1y = 3(1) - 2 = 1

2Problem 2medium

Question:

Where does g(x)=x33x2g(x) = x^3 - 3x^2 have horizontal tangent lines?

💡 Show Solution

Horizontal tangent lines occur where g(x)=0g'(x) = 0.

Step 1: Find the derivative g(x)=3x26xg'(x) = 3x^2 - 6x

Step 2: Set equal to zero 3x26x=03x^2 - 6x = 0

Step 3: Factor 3x(x2)=03x(x - 2) = 0

Step 4: Solve x=0 or x=2x = 0 \text{ or } x = 2

Step 5: Find the y-coordinates

  • At x=0x = 0: g(0)=00=0g(0) = 0 - 0 = 0
  • At x=2x = 2: g(2)=812=4g(2) = 8 - 12 = -4

Answer: Horizontal tangent lines at (0,0)(0, 0) and (2,4)(2, -4)

These are the critical points where the function changes from increasing to decreasing or vice versa!