Derivatives and Tangent Lines: Calculus Study Notes

Derivatives and Tangent Lines

Introduction

The concept of the derivative is fundamental in calculus, representing the instantaneous rate of change of a function. This study guide covers how to estimate slopes from graphs, calculate derivatives using the definition, and find equations of tangent lines to curves.

Estimating the Slope of a Curve at a Point

Understanding Slope and Tangency

• Slope at a Point: The slope of a curve at a specific point is the slope of the tangent line to the curve at that point.

• Tangent Line: A straight line that touches a curve at a single point without crossing it locally.

• Undefined Slope: If the tangent is vertical, the slope is undefined.

Example: If a curve has a sharp corner or a vertical tangent at a point, the slope at that point is undefined.

The Derivative: Definition and Calculation

Definition of the Derivative

• The derivative of a function f(x) at a point x = a is defined as:

$f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$

• This limit, if it exists, gives the instantaneous rate of change of f at a.

Calculating the Derivative Using the Definition

• Given a function, substitute into the definition and simplify.

• Example: For $f(x) = 5x^2 + 7$, the derivative is:

$f'(x) = \lim_{h \to 0} \frac{5(x+h)^2 + 7 - [5x^2 + 7]}{h}$

$= \lim_{h \to 0} \frac{5(x^2 + 2xh + h^2) + 7 - 5x^2 - 7}{h}$

$= \lim_{h \to 0} \frac{5x^2 + 10xh + 5h^2 + 7 - 5x^2 - 7}{h}$

$= \lim_{h \to 0} \frac{10xh + 5h^2}{h}$

$= \lim_{h \to 0} (10x + 5h) = 10x$

• Thus, $f'(x) = 10x$.

Evaluating the Derivative at Specific Points

• To find $f'(0)$, substitute $x = 0$ into the derivative:

$f'(0) = 10 \times 0 = 0$

• To find $f'(9)$, substitute $x = 9$:

$f'(9) = 10 \times 9 = 90$

Equation of the Tangent Line

Finding the Equation of a Tangent Line

• The equation of the tangent line to the curve $y = f(x)$ at the point $(a, f(a))$ is:

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

• Here, $f'(a)$ is the slope of the tangent at $x = a$.

Examples

• For $y = x^2 - 4$ at $(-2, 0)$:

$f(x) = x^2 - 4$ $f' (x) = 2x$ $f'(-2) = 2(-2) = -4$ Equation: $y - 0 = -4(x + 2)$ $y = -4x - 8$

• For $y = x^2 + 1$ at $(2, 5)$:

$f(x) = x^2 + 1$ $f'(x) = 2x$ $f'(2) = 4$ Equation: $y - 5 = 4(x - 2)$ $y = 4x - 3$

Summary Table: Derivative and Tangent Line Formulas

Function	Derivative $f'(x)$	Tangent Line at $(a, f(a))$
$x^2 - 4$	$2x$	$y = 2a(x - a) + f(a)$
$x^2 + 1$	$2x$	$y = 2a(x - a) + f(a)$
$5x^2 + 7$	$10x$	$y = 10a(x - a) + f(a)$

Key Points to Remember

• The slope of a curve at a point is the value of the derivative at that point.

• If the tangent is vertical or the function is not differentiable at a point, the slope is undefined.

• The equation of the tangent line uses the point-slope form with the derivative as the slope.

• Always use the definition of the derivative for rigorous calculation, especially when first learning the concept.