The Derivative Function
Textbook Reference
| Primary source | OpenStax Calculus Volume 1, Section 3.2: “The Derivative as a Function” |
| Book URL | https://openstax.org/details/books/calculus-volume-1 |
Freely available and openly licensed.
Key idea
In the previous lesson, you computed $f'(a)$ -- the derivative at a single point $a$. Now treat $a$ as a variable and write $x$ in its place. The result is $f'(x)$: a new function whose output at each $x$ is the slope of $f$’s graph at that $x$.
This is a conceptual shift: $f'$ is not a number at one point. It is a function that assigns a slope to every point in $f$’s domain where the derivative exists. Computing $f'(x)$ from the definition -- letting $a$ be any $x$ and simplifying -- produces formulas like $f'(x) = 2x$ (for $f(x) = x^2$), which then let you find the derivative at any point with a single substitution.
Prerequisite Check
Before this lesson, make sure you can do all of the following:
Quick Reference
The derivative function: \[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}. \]
Computing from the definition. Treat $x$ as a fixed but arbitrary point. Expand $f(x+h)$, subtract $f(x)$, divide by $h$, cancel $h$ from the numerator, and take $\lim_{h \to 0}$.
Notations for the derivative function: $f'(x)$, $\dfrac{dy}{dx}$, $\dfrac{df}{dx}$, $y'$, $Df$.
Key Concepts
1. Deriving $f'(x)$ for Polynomials
Example 1. Find $f'(x)$ for $f(x) = x^2$.
$f(x+h) = (x+h)^2 = x^2 + 2xh + h^2$.
$f(x+h) - f(x) = 2xh + h^2 = h(2x + h)$.
$\dfrac{f(x+h)-f(x)}{h} = 2x + h$.
$f'(x) = \lim_{h \to 0}(2x + h) = 2x$.
So $f'(x) = 2x$. The slope of $y = x^2$ at any point $x$ is $2x$.
- At $x = 3$: slope $= 6$. At $x = -1$: slope $= -2$. At $x = 0$: slope $= 0$ (horizontal tangent at the vertex).
Example 2. Find $f'(x)$ for $f(x) = x^3$.
$(x+h)^3 = x^3 + 3x^2h + 3xh^2 + h^3$.
$f(x+h) - f(x) = 3x^2h + 3xh^2 + h^3 = h(3x^2 + 3xh + h^2)$.
$f'(x) = \lim_{h \to 0}(3x^2 + 3xh + h^2) = 3x^2$.
Example 3. Find $f'(x)$ for $f(x) = mx + b$ (a linear function).
$f(x+h) = m(x+h) + b = mx + mh + b$.
$f(x+h) - f(x) = mh$.
$\dfrac{mh}{h} = m$.
$f'(x) = m$. The derivative of a linear function is its slope, which is constant.
2. Deriving $f'(x)$ for Rational Functions
Example 4. Find $f'(x)$ for $f(x) = \dfrac{1}{x}$.
$f(x+h) = \dfrac{1}{x+h}$.
$f(x+h) - f(x) = \dfrac{1}{x+h} - \dfrac{1}{x} = \dfrac{x - (x+h)}{x(x+h)} = \dfrac{-h}{x(x+h)}$.
$\dfrac{f(x+h)-f(x)}{h} = \dfrac{-1}{x(x+h)}$.
$f'(x) = \lim_{h \to 0} \dfrac{-1}{x(x+h)} = \dfrac{-1}{x^2}$.
So $f'(x) = -\dfrac{1}{x^2}$ for $x \neq 0$.
3. The Derivative as Slope Information
Once you have $f'(x)$, you can read off geometric behavior:
- Where $f'(x) > 0$: $f$ is increasing (graph rises left to right).
- Where $f'(x) < 0$: $f$ is decreasing (graph falls left to right).
- Where $f'(x) = 0$: $f$ has a horizontal tangent (flat; potential max or min).
Example 5. For $f(x) = x^2$, $f'(x) = 2x$.
$f'(x) > 0$ when $x > 0$: $f$ is increasing for $x > 0$.
$f'(x) < 0$ when $x < 0$: $f$ is decreasing for $x < 0$.
$f'(0) = 0$: horizontal tangent at the vertex $(0, 0)$.
This matches the parabola: it falls to the left of the origin and rises to the right.
4. Notation
The notation $\dfrac{dy}{dx}$ (read “dee $y$ dee $x$”) means the derivative of $y$ with respect to $x$. When $y = f(x)$, it equals $f'(x)$.
The notation emphasizes the quotient $\Delta y / \Delta x$ that the derivative is the limit of. The $d$ replaces $\Delta$ to signal infinitesimal (limiting) differences.
At a specific point $x = a$: \[ \left.\frac{dy}{dx}\right|_{x=a} = f'(a). \]
All the following mean the same thing: $f'(x)$, $y'$, $\dfrac{dy}{dx}$, $\dfrac{d}{dx}[f(x)]$.
Common Errors
| Error | Example | Correction |
|---|---|---|
| Treating $x$ as constant while computing | Writing $f(x+h) = f(x) + f(h)$ (which is false in general) | $f(x+h)$ means substitute $x+h$ everywhere; expand carefully |
| Carrying $h$ into the answer | $f'(x) = 2x + h$ (not taking the limit) | Take $\lim_{h \to 0}$; $h \to 0$, so $h$ vanishes |
| Confusing $f'(x)$ with $f(x)$ | Writing $f'(x) = x^2$ when $f(x) = x^2$ | The derivative of $x^2$ is $2x$, not $x^2$ |
| Forgetting to expand $(x+h)^n$ completely | $(x+h)^2 = x^2 + h^2$ (missing $2xh$) | $(x+h)^2 = x^2 + 2xh + h^2$; never omit the middle term |
Leveled Practice
Level 1 -- Finding $f'(x)$ from the Definition
Problem 1. Find $f'(x)$ for $f(x) = 4x + 7$.
Show answer
$f(x+h) - f(x) = 4h$. Quotient: $4$. $f'(x) = 4$. (Constant slope of a linear function.)
Problem 2. Find $f'(x)$ for $f(x) = 5x^2 - 3$.
Show answer
$f(x+h) = 5(x+h)^2 - 3 = 5x^2 + 10xh + 5h^2 - 3$.
$f(x+h) - f(x) = 10xh + 5h^2 = h(10x + 5h)$.
$f'(x) = \lim_{h \to 0}(10x + 5h) = 10x$.
Problem 3. Find $f'(x)$ for $f(x) = \dfrac{1}{x + 1}$.
Show answer
$f(x+h) - f(x) = \dfrac{1}{x+h+1} - \dfrac{1}{x+1} = \dfrac{(x+1)-(x+h+1)}{(x+h+1)(x+1)} = \dfrac{-h}{(x+h+1)(x+1)}$.
$\dfrac{f(x+h)-f(x)}{h} = \dfrac{-1}{(x+h+1)(x+1)}$.
$f'(x) = \dfrac{-1}{(x+1)^2}$.
Level 2 -- Interpreting $f'(x)$
Problem 4. For $f(x) = x^2 - 4x$, find $f'(x)$ and determine where $f$ is increasing, decreasing, and has a horizontal tangent.
Show answer
$f(x+h) = (x+h)^2 - 4(x+h) = x^2 + 2xh + h^2 - 4x - 4h$.
$f(x+h) - f(x) = 2xh + h^2 - 4h = h(2x + h - 4)$.
$f'(x) = 2x - 4$.
$f'(x) > 0$ when $x > 2$: increasing. $f'(x) < 0$ when $x < 2$: decreasing. $f'(2) = 0$: horizontal tangent at $x = 2$.
Level 3 -- Deriving a Root Function
Problem 5. Find $f'(x)$ for $f(x) = \sqrt{x}$ ($x > 0$) using the definition and conjugate multiplication.
Show answer
$f(x+h) - f(x) = \sqrt{x+h} - \sqrt{x}$.
Multiply by conjugate:
$\dfrac{\sqrt{x+h}-\sqrt{x}}{h} \cdot \dfrac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}} = \dfrac{(x+h)-x}{h(\sqrt{x+h}+\sqrt{x})} = \dfrac{1}{\sqrt{x+h}+\sqrt{x}}$.
$f'(x) = \lim_{h \to 0} \dfrac{1}{\sqrt{x+h}+\sqrt{x}} = \dfrac{1}{2\sqrt{x}}$.
Common Misconceptions
$f'(x)$ and $f(x)$ measure the same thing at each point. The height $f(x)$ describes where the graph sits above the axis. The derivative $f'(x)$ describes how steeply the graph is rising or falling there. For $f(x) = x^2$, the values $f(3) = 9$ and $f'(3) = 6$ are different numbers measuring different things. A function can be large in value while nearly flat (slope close to zero), or small in value while very steep.
$f'(x)$ is a single number computed once and applied everywhere. Students sometimes compute $f'(a)$ for one value of $a$ and then use that number for all $x$. The derivative $f'(x)$ is a function: it produces a different slope value at each input. For $f(x) = x^2$, $f'(1) = 2$ and $f'(3) = 6$; the slope changes as $x$ changes, and no single number describes the slope everywhere.
Mastery Checklist
You have mastered this skill when you can do all of the following without referring to notes:
Mental Model
Think of $f'(x)$ as the “slope at each $x$” function. While $f(x)$ tells you the height of the graph at each $x$, the function $f'(x)$ tells you the tilt of the graph at each $x$.
Where $f'(x)$ is large and positive, $f$ is rising steeply. Where $f'(x)$ is near zero, $f$ is nearly flat. Where $f'(x)$ is large and negative, $f$ is falling steeply. The graph of $f'$ encodes the entire geometry of $f$’s slope.
This is why reading the graph of $f'$ tells you so much: peaks and valleys of $f$ occur at zeros of $f'$; steep regions of $f$ correspond to regions where $|f'|$ is large.
Connections
Within Calculus I (MATH161)
- Power rule: The definition calculations above produce $d(x^n)/dx = nx^{n-1}$; the power rule is the general pattern confirmed by these specific cases.
- Higher derivatives: Once $f'(x)$ is a function, you can differentiate it again to get $f''(x)$ (second derivative), which measures the rate of change of the slope.
- Graphical analysis: The sign of $f'$ tells you where $f$ is increasing or decreasing, the foundation of all curve-sketching and optimization.
Back to Calculus I Skills | Previous: The Derivative at a Point | Next: Differentiability