Chapter 4: Definition of the Derivative
The Derivative at a Point and the Derivative Function
4.1 The Derivative at a Point
Building on the groundwork laid so far, we now give the formal definition of the derivative.
The derivative of a function $f(x)$ at $x = a$ is defined by the following limit:
$$f'(a) = \lim_{h \to 0} \dfrac{f(a+h) - f(a)}{h}$$Here $h$ approaches $0$ from both the positive and the negative side. If it approaches the same value from both sides, that is, if this limit exists, we say that $f(x)$ is differentiable at $x = a$.
As shown in Figure 1, the derivative $f'(a)$ represents the slope of the tangent line to the curve $y = f(x)$ at the point $(a, f(a))$.
Alternative Form
By substituting $x = a + h$ (so that $x \to a$ as $h \to 0$), the derivative can also be written as
$$f'(a) = \lim_{x \to a} \dfrac{f(x) - f(a)}{x - a}$$4.2 The Derivative Function
Instead of computing the derivative at a single fixed point $a$, we define the derivative function as the function that gives the derivative $f'(x)$ at the variable $x$.
The derivative function of $f(x)$ is:
$$f'(x) = \lim_{h \to 0} \dfrac{f(x+h) - f(x)}{h}$$The derivative function $f'(x)$ is a new function that gives the "slope of the tangent = rate of change" at each point $x$ of the original function $f(x)$. The larger the absolute value of the derivative, the more $f(x)$ changes when $x$ is moved slightly.
- If $x=t$ represents time and $f(t)$ represents position, then $f'(t)$ is velocity.
- If $f(x)$ is "quantity," then $f'(x)$ is "rate of increase"
- If $f(x)$ is a graph, then $f'(x)$ is "the slope at each point"
4.3 Notation for Derivatives
When $y=f(x)$, the notations $f'(x)$, $df/dx$, and $dy/dx$ can represent the same derivative function. They are different notational conventions and are used in slightly different contexts.
| Notation | Read as | Notes |
|---|---|---|
| $f'(x)$ | "f prime of x" | Lagrange's notation. The most common. |
| $\dfrac{df}{dx}$ | "dee f dee x" | Leibniz's notation. Can be manipulated like a fraction. |
| $\dfrac{dy}{dx}$ | "dee y dee x" | Used when $y = f(x)$. Makes the rate-of-change meaning explicit. |
| $\dfrac{d}{dx}f(x)$ | "dee dee x of f of x" | Treats $\dfrac{d}{dx}$ as a "differentiation operator." |
| $\dot{y}$ | "y dot" | Newton's notation. Used for time derivatives. |
Let $y=f(x)$ and define $\Delta y=f(x+\Delta x)-f(x)$. When the derivative exists,
$$\dfrac{dy}{dx}=\lim_{\Delta x\to0}\dfrac{\Delta y}{\Delta x}$$a single symbol representing the entire limit operation. While $\dfrac{\Delta y}{\Delta x}$ is indeed a fraction (a ratio of finite changes), after taking the limit $\Delta x \to 0$ $\dfrac{dy}{dx}$ denotes the resulting limit value. (At university, you will also learn a way of giving $dy$ and $dx$ their own meanings so that $dy/dx$ can be treated as a ratio.)
However, in the chain rule for composite functions,
$$\dfrac{dy}{dx} = \dfrac{dy}{du} \cdot \dfrac{du}{dx}$$it looks as if $du$ cancels out. In integration by substitution, too, one can write $dx = \dfrac{dx}{du}\,du$ as if $dx$ were a fraction. The fact that the rules of fractions often carry over to computations with $dy/dx$ is a major advantage of Leibniz notation.
4.4 Differentiability
Not every function is differentiable.
A function $f(x)$ is said to be differentiable at $x = a$ if
$$\lim_{h \to 0} \dfrac{f(a+h) - f(a)}{h}$$exists (i.e., is a finite number).
A Non-Differentiable Example: A Corner
Consider $f(x) = |x|$. Let us examine the derivative at $x = 0$.
Left-hand limit (when $h < 0$):
$$\lim_{h \to 0^-} \dfrac{|0+h| - |0|}{h} = \lim_{h \to 0^-} \dfrac{|h|}{h} = \lim_{h \to 0^-} \dfrac{-h}{h} = -1$$Right-hand limit (when $h > 0$):
$$\lim_{h \to 0^+} \dfrac{|0+h| - |0|}{h} = \lim_{h \to 0^+} \dfrac{|h|}{h} = \lim_{h \to 0^+} \dfrac{h}{h} = +1$$Since the left-hand limit $(-1)$ and the right-hand limit $(+1)$ disagree, the limit does not exist.
Therefore, $f(x) = |x|$ is not differentiable at $x = 0$.
If $f(x)$ is differentiable at $x = a$, then $f(x)$ is continuous at $x = a$.
(The converse is false: a function can be continuous without being differentiable.)
4.5 Worked Examples
Example 1: Constant Function $f(x) = c$
Step 1: Set up the definition
$$f'(x) = \lim_{h \to 0} \dfrac{f(x+h) - f(x)}{h}$$Step 2: Substitute $f(x+h)$ and $f(x)$
Since $f(x)$ is constant, $f(x) = f(x+h) = c$. Therefore:
$$f'(x) = \lim_{h \to 0} \dfrac{c - c}{h} = \lim_{h \to 0} \dfrac{0}{h} = \lim_{h \to 0} 0 = 0$$Result: $f'(x) = 0$
Example 2: Linear Function $f(x) = mx + n$
Step 1: Compute $f(x+h)$
$$f(x+h) = m(x+h) + n = mx + mh + n$$Step 2: Compute $f(x+h) - f(x)$
$$f(x+h) - f(x) = (mx + mh + n) - (mx + n) = mh$$Step 3: Divide by $h$
$$\dfrac{f(x+h) - f(x)}{h} = \dfrac{mh}{h} = m \quad (h \neq 0)$$Step 4: Take the limit
$$f'(x) = \lim_{h \to 0} m = m$$Result: $f'(x) = m$
The derivative of a linear function equals its slope $m$, which matches our intuition.
Example 3: Quadratic Function $f(x) = x^2$
Step 1: Expand $f(x+h)$
$$f(x+h) = (x+h)^2 = x^2 + 2xh + h^2$$Step 2: Compute $f(x+h) - f(x)$
$$f(x+h) - f(x) = (x^2 + 2xh + h^2) - x^2 = 2xh + h^2$$Step 3: Divide by $h$
$$\dfrac{f(x+h) - f(x)}{h} = \dfrac{2xh + h^2}{h} = \dfrac{h(2x + h)}{h} = 2x + h \quad (h \neq 0)$$Step 4: Take the limit
$$f'(x) = \lim_{h \to 0} (2x + h) = 2x$$Result: $(x^2)' = 2x$
Chapter Summary
- Derivative at a point: $f'(a) = \displaystyle\lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$ (a constant representing the slope of the tangent at $a$)
- Derivative function: $f'(x) = \displaystyle\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$ (a function giving the slope at each point $x$)
- Notation: $f'(x)$, $\dfrac{df}{dx}$, $\dfrac{dy}{dx}$, etc.
- Differentiable: the limit exists
- Basic formulas: $(c)' = 0$, $(mx + n)' = m$, $(x^2)' = 2x$
Exercises
Problem 1 (Basic): For $f(x) = 3x - 2$, find $f'(x)$ from the definition of the derivative.
Hint
$\lim_{h \to 0} \dfrac{(3(x+h)-2)-(3x-2)}{h} = \lim_{h \to 0} \dfrac{3h}{h} = 3$.
Problem 2 (Standard): For $f(x) = x^2 + 2x$, find $f'(x)$ from the definition of the derivative.
Hint
Expand $(x+h)^2 + 2(x+h) = x^2 + 2xh + h^2 + 2x + 2h$. After simplifying, letting $h \to 0$ gives $2x + 2$.
Problem 3 (Applied): For $f(x) = \dfrac{1}{x}$ ($x\ne0$), find $f'(x)$ from the definition of the derivative.
Hint
$\dfrac{f(x+h)-f(x)}{h} = \dfrac{1}{h}\left(\dfrac{1}{x+h} - \dfrac{1}{x}\right) = \dfrac{-h}{h \cdot x(x+h)} \to -\dfrac{1}{x^2}$.