PROJECTENGRAM

THE TANGENT LINE AND THE DERIVATIVE

The Tangent Line and the Derivative

Tags: #◇Computer Science #◇CSMATH1 #◇Problem Solving

◇Overview

The Tangent Line

A tangent line is a straight line that "touches" a given curve at a specific point (the point of tangency). The slope of the tangent line matches the slope of the curve at that exact point. To find the slope of the tangent line, we consider a secant line passing through two points on the curve: (x,f(x))(x, f(x)) and (x+h,f(x+h))(x+h, f(x+h)).
The slope of this secant line is given by the difference quotient:
ΔyΔx=f(x+h)−f(x)h\frac{\Delta y}{\Delta x} = \frac{f(x+h) - f(x)}{h}
As the distance hh between the points approaches 0, the secant line approaches the tangent line.

The Derivative

The derivative of a function ff with respect to xx provides the slope of the tangent line to the curve at any point (x,f(x))(x, f(x)). It is also interpreted as the instantaneous rate of change of the function.
The derivative is denoted by f′(x)f'(x), dydx\frac{dy}{dx}, ddxf(x)\frac{d}{dx}f(x), or DxyD_x y, and is defined by the limit:
f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}

Curves without Tangent Lines

The tangent line to a graph at a point may not exist. This occurs in two main scenarios:

  1. Discontinuity: The function is not continuous at the point.
  2. Sharp Corners or Cusps: The function's graph has a sharp turn, meaning the limit from the left and the limit from the right yield different slopes.
    (Note: A vertical tangent line is allowed in the definition; it occurs when the slope approaches ±∞\pm \infty.)

◇Important Formulas

Difference Quotient (Secant Slope):
msec=f(x+h)−f(x)hm_{\text{sec}} = \frac{f(x+h) - f(x)}{h}
Limit Definition of the Derivative:
f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}

◇Worked Examples

Example 1: Finding the derivative using the limit definition
Find f′(x)f'(x) for f(x)=−x2f(x) = -x^2.
Solution:
f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}
f′(x)=lim⁡h→0−(x+h)2−(−x2)hf'(x) = \lim_{h \to 0} \frac{-(x+h)^2 - (-x^2)}{h}
f′(x)=lim⁡h→0−(x2+2xh+h2)+x2hf'(x) = \lim_{h \to 0} \frac{-(x^2 + 2xh + h^2) + x^2}{h}
f′(x)=lim⁡h→0−2xh−h2hf'(x) = \lim_{h \to 0} \frac{-2xh - h^2}{h}
f′(x)=lim⁡h→0−h(2x+h)h=lim⁡h→0−(2x+h)=−2xf'(x) = \lim_{h \to 0} \frac{-h(2x + h)}{h} = \lim_{h \to 0} -(2x + h) = -2x

Example 2: Finding the slope of a perpendicular line
Find the slope of the line perpendicular to the tangent of the graph of f(x)=−x2f(x) = -x^2 at x=2x = 2.
Solution:
From Example 1, f′(x)=−2xf'(x) = -2x.
The slope of the tangent line at x=2x=2 is mtan=f′(2)=−2(2)=−4m_{\text{tan}} = f'(2) = -2(2) = -4.
The perpendicular (or normal) line has a slope that is the negative reciprocal of the tangent's slope.
mperp=−1mtan=−1−4=14m_{\text{perp}} = -\frac{1}{m_{\text{tan}}} = -\frac{1}{-4} = \frac{1}{4}

◇References & Navigation

CONFIRMATION // 理解度の確認

ACTIVE RECALL VERIFICATION

Select an answer to verify conceptual comprehension and reveal the technical rationale.

Q1

Which limit expression defines the derivative f′(x)f'(x) of a function ff at xx?

Q2

Geometrically, what does the difference quotient f(a+h)−f(a)h\frac{f(a + h) - f(a)}{h} represent before taking the limit as h→0h \to 0?

Q3

Find the equation of the tangent line to the parabola f(x)=x2+1f(x) = x^2 + 1 at the point (2,5)(2, 5).

Q4

If s(t)s(t) gives the position of a particle along a coordinate line at time tt, what is the physical interpretation of the derivative s′(t)s'(t)?

Q5

What is the relationship between the slope of the tangent line mT=f′(a)m_T = f'(a) (assuming f′(a)≠0f'(a) \ne 0) and the slope of the normal line mNm_N at the same point?

The Tangent Line and the Derivative | PROJECT ENGRAM