Lesson 5: The Derivative
Secant and Tangent Lines
Before introducing the derivative, we must understand two fundamental types of lines associated with functions: secant lines and tangent lines. These concepts provide the geometric foundation for understanding derivatives.
Definition: A secant line of a function \(f(x)\) is a line that passes through two distinct points of \(f(x)\).
Secant lines represent the average rate of change of a function between two points. If we have two points \((x_1, f(x_1))\) and \((x_2, f(x_2))\) on the graph of \(f(x)\), the secant line connects these points and has slope equal to the average rate of change over that interval.
Definition: The tangent line to a function \(f(x)\) at the point \(x = c\) is the line that touches the graph of \(f(x)\) at the point \((c, f(c))\). We can think of this as taking the slope of the graph at the point \((c, f(c))\) and extending it into a line with the same slope.
Important: The slope of the tangent line to \(f(x)\) at \(x = c\) is the slope of the graph of \(f(x)\) at the point \((c, f(c))\).
The tangent line represents the instantaneous rate of change of the function at a specific point. Unlike the secant line which connects two points, the tangent line just touches the curve at a single point and indicates the direction the function is heading at that exact location.
Finding the Slope of the Tangent Line Using Slopes of Secant Lines
The key insight in calculus is that we can find the slope of a tangent line by examining secant lines and letting the two points get closer and closer together. This limit process transforms the average rate of change into an instantaneous rate of change.
Diagram Description: The diagram shows a curve representing \(f(x)\) with a point labeled \((x, f(x))\) and another point to its right labeled \((x+h, f(x+h))\). A secant line passes through both points. The horizontal distance between the points is labeled \(h\), and the vertical distance is labeled \(f(x+h) - f(x)\). This illustrates how the secant line connects two points on the curve separated by a horizontal distance of \(h\).
The slope of the secant line shown in the diagram is calculated using the slope formula:
\[\text{Slope of secant line} = \frac{f(x+h) - f(x)}{h}\]This formula gives the average rate of change of \(f(x)\) over the interval from \(x\) to \(x+h\). We call it the difference quotient. As we make \(h\) smaller and smaller, the secant line approaches the tangent line, and the average rate of change approaches the instantaneous rate of change.
To find the slope of the tangent line, we take the limit as \(h\) approaches 0:
\[\text{Slope of tangent line} = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\]This limit, when it exists, gives us the exact slope of the curve at the point \((x, f(x))\).
The Derivative
Definition: The derivative of a function \(f(x)\) at the point \(x\) is the slope of the tangent line to \(f(x)\) at the point \(x\). We write
\[f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\]The derivative \(f'(x)\) is itself a function that tells us the slope of the tangent line to \(f(x)\) at any value of \(x\). When we evaluate \(f'(c)\) for a specific value \(c\), we get the slope of the tangent line at that particular point.
Notation
Notation: If \(y = f(x)\), then we can write the derivative using any of the following equivalent notations:
\[y' = f'(x) = \frac{dy}{dx} = \frac{d}{dx}f(x)\]The notation \(f'(x)\) is called Lagrange notation, while \(\frac{dy}{dx}\) and \(\frac{d}{dx}f(x)\) are called Leibniz notation. The Leibniz notation emphasizes that the derivative is a limit of ratios of differences, while the prime notation is more compact for computation.
Geometric Interpretation
Geometric Interpretation: The slope of the tangent line to \(f(x)\) at the point \(x = c\) is equal to the derivative of \(f(x)\) at the point \(x = c\). That is, \(f'(c)\) is the slope of the tangent line to \(f(x)\) at \(x = c\).
This geometric interpretation connects the algebraic definition of the derivative (as a limit) with its visual meaning (as the slope of a tangent line). It allows us to interpret derivatives in terms of rates of change and provides intuition for understanding what derivatives tell us about functions.
Examples
Example 1: Derivative of a Linear Function
Suppose \(f(x) = x + 5\). Find \(f'(x)\).
Solution:
We use the limit definition of the derivative:
\[f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\]First, we find \(f(x+h)\):
\[f(x+h) = (x+h) + 5 = x + h + 5\]Now we substitute into the limit definition:
\[f'(x) = \lim_{h \to 0} \frac{(x + h + 5) - (x + 5)}{h}\]Simplify the numerator:
\[f'(x) = \lim_{h \to 0} \frac{x + h + 5 - x - 5}{h} = \lim_{h \to 0} \frac{h}{h}\]Cancel the \(h\) (valid since \(h \neq 0\) in the limit):
\[f'(x) = \lim_{h \to 0} 1 = 1\]Therefore, \(f'(x) = 1\).
This result makes sense: the function \(f(x) = x + 5\) is a line with slope 1, so its derivative (the slope of the tangent line) is constantly 1 everywhere.
Example 2: Derivative of a Quadratic Function
Suppose \(f(x) = 3x^2 + 1\). Find \(f'(x)\).
Solution:
We use the limit definition of the derivative:
\[f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\]First, we find \(f(x+h)\):
\[f(x+h) = 3(x+h)^2 + 1 = 3(x^2 + 2xh + h^2) + 1 = 3x^2 + 6xh + 3h^2 + 1\]Now we substitute into the limit definition:
\[f'(x) = \lim_{h \to 0} \frac{(3x^2 + 6xh + 3h^2 + 1) - (3x^2 + 1)}{h}\]Simplify the numerator:
\[f'(x) = \lim_{h \to 0} \frac{3x^2 + 6xh + 3h^2 + 1 - 3x^2 - 1}{h} = \lim_{h \to 0} \frac{6xh + 3h^2}{h}\]Factor out \(h\) from the numerator:
\[f'(x) = \lim_{h \to 0} \frac{h(6x + 3h)}{h}\]Cancel the \(h\) (valid since \(h \neq 0\) in the limit):
\[f'(x) = \lim_{h \to 0} (6x + 3h)\]Take the limit as \(h\) approaches 0:
\[f'(x) = 6x + 3\cdot 0 = 6x\]Therefore, \(f'(x) = 6x\).
Example 3: Finding the Equation of a Tangent Line
Find the equation of the tangent line to the graph of \(f(x) = 3x^2 + 1\) at the point \(x = 2\).
Solution:
To find the equation of the tangent line, we need a point and a slope. We already know the point of tangency occurs at \(x = 2\), and we found in example 2 above that \(f'(x) = 6x\).
First, find the \(y\)-coordinate of the point of tangency:
\[f(2) = 3\cdot 2^2 + 1 = 3\cdot 4 + 1 = 12 + 1 = 13\]So, the point of tangency is \((2, 13)\).
Next, find the slope of the tangent line at \(x = 2\):
\[f'(2) = 6\cdot 2 = 12\]Now, we use the point-slope form of a line: \(y - y_1 = m(x - x_1)\), where \(m\) is the slope and \((x_1, y_1)\) is the point.
\[y - 13 = 12(x - 2)\]Simplify to get the equation in slope-intercept form:
\[y - 13 = 12x - 24\] \[y = 12x - 24 + 13\] \[y = 12x - 11\]Therefore, the equation of the tangent line is \(y = 12x - 11\).
Example 4: Working Backwards from the Derivative
The derivative of a function \(f(x)\) is found by computing
\[f'(x) = \lim_{h \to 0} \frac{\sqrt{(x+h)^3 + 2} - \sqrt{x^3 + 2}}{h}\]What is \(f(x)\)?
Solution:
Recall that the definition of the derivative is:
\[f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\]Comparing this with the given expression:
\[f'(x) = \lim_{h \to 0} \frac{\sqrt{(x+h)^3 + 2} - \sqrt{x^3 + 2}}{h}\]We can identify that:
\[f(x+h) = \sqrt{(x+h)^3 + 2}\] \[f(x) = \sqrt{x^3 + 2}\]Therefore, \(f(x) = \sqrt{x^3 + 2}\).
Key Takeaway: The derivative is defined as the limit of the difference quotient and represents the instantaneous rate of change of a function. Geometrically, it gives the slope of the tangent line at any point on the curve. To find derivatives from the definition, we must carefully substitute, simplify, factor, and then take the limit as \(h\) approaches 0.