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The Derivative Definition

Reference: Stewart 2.7  •  Chapter: 2  •  Section: 7

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Why Do We Need a New Limit?

You've learned how to compute limits. Now here's a question: how steep is a curve at a single point?

A straight line has constant slope everywhere. But for curves like $y = x^2$, the steepness changes as you move along it. Near $x = 0$, the parabola is almost flat. Near $x = 3$, it's quite steep. How do we capture this "instantaneous steepness" mathematically?

The answer is a special kind of limit called the derivative. It measures the slope at exactly one point by "zooming in" until the curve looks like a straight line.

Prerequisite Map

PrerequisitesFinding Tangent Lines Using the DerivativeInstantaneous Velocity from the DerivativeBasic Limit Laws
This skillThe Derivative Definition
Leads toFinding Tangent Lines Using the DerivativeInstantaneous Velocity from the Derivative

Quick Reference

Property Value
Section Stewart 2.1
Course MATH161
Difficulty Intermediate
Time ~20 minutes

Key Concepts

The Difference Quotient

Before we can define the derivative, we need the difference quotient:

$$\frac{f(a+h) - f(a)}{h}$$

This measures the average rate of change of $f$ over the interval from $x = a$ to $x = a + h$.

                     Q = (a+h, f(a+h))
                    /
                   /  ← Secant line (slope = difference quotient)
                  /
        ─────────●─────────
                /
               / P = (a, f(a))
              /

The Two Forms of the Derivative

The derivative of $f$ at the point $x = a$ is defined as:

Form 1 (h-form): $$\boxed{f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}}$$

Form 2 (x-form): $$\boxed{f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a}}$$

Both forms give the same answer. The h-form is often easier for computation; the x-form shows clearly that we're taking a limit as a second point approaches our point.

Why These Forms Are Equivalent

If we let $x = a + h$, then:

So Form 1 transforms into Form 2 directly.

The Computation Process

To find $f'(a)$ using the definition:

Step 1: Write out the difference quotient $\frac{f(a+h) - f(a)}{h}$

Step 2: Substitute and expand $f(a+h)$

Step 3: Simplify algebraically (factor out $h$)

Step 4: Cancel the $h$ in numerator and denominator

Step 5: Evaluate the limit as $h \to 0$

Worked Example: Finding $f'(2)$ for $f(x) = x^2 - 3x$

Step 1: Set up the difference quotient: $$f'(2) = \lim_{h \to 0} \frac{f(2+h) - f(2)}{h}$$

Step 2: Compute $f(2+h)$ and $f(2)$:

Step 3: Form the quotient: $$\frac{(h^2 + h - 2) - (-2)}{h} = \frac{h^2 + h}{h}$$

Step 4: Factor and cancel: $$\frac{h(h + 1)}{h} = h + 1$$

Step 5: Take the limit: $$f'(2) = \lim_{h \to 0}(h + 1) = 1$$

So the slope of $f(x) = x^2 - 3x$ at $x = 2$ is exactly $1$.

Practice Problems

Level 1 Recognizing the Derivative Definition

Which of the following limits represents the derivative of $f(x) = x^3$ at $x = 2$?

(A) $\displaystyle\lim_{h \to 0} \frac{(2+h)^3 - 8}{h}$

(B) $\displaystyle\lim_{x \to 2} \frac{x^3 - 2}{x - 2}$

(C) $\displaystyle\lim_{h \to 0} \frac{h^3 - 8}{h}$

(D) $\displaystyle\lim_{x \to 0} \frac{x^3 - 8}{x - 2}$

Thought Process

Compare each option to the two definition forms:

  • Form 1: $f'(2) = \lim_{h \to 0} \frac{f(2+h) - f(2)}{h}$
  • Form 2: $f'(2) = \lim_{x \to 2} \frac{f(x) - f(2)}{x - 2}$

For $f(x) = x^3$: $f(2) = 8$ and $f(2+h) = (2+h)^3$.

Check which option matches one of these forms.

Show Answer

(A) is correct.

Using Form 1: $f'(2) = \lim_{h \to 0} \frac{f(2+h) - f(2)}{h} = \lim_{h \to 0} \frac{(2+h)^3 - 8}{h}$

Option (B) has $f(2) = 2$ instead of $8$. Option (C) has $f(h)$ instead of $f(2+h)$. Option (D) takes the limit as $x \to 0$ instead of $x \to 2$.

Level 2 Direct Application of the Definition

Use the limit definition to find $f'(3)$ for $f(x) = 2x^2 + 1$.

Thought Process

Apply the h-form definition:

  1. Write $\frac{f(3+h) - f(3)}{h}$
  2. Compute $f(3+h) = 2(3+h)^2 + 1$
  3. Compute $f(3) = 2(9) + 1 = 19$
  4. Simplify the quotient until you can cancel $h$
  5. Take the limit as $h \to 0$
Show Answer

$$f'(3) = \lim_{h \to 0} \frac{f(3+h) - f(3)}{h}$$

Compute $f(3+h)$: $$f(3+h) = 2(3+h)^2 + 1 = 2(9 + 6h + h^2) + 1 = 18 + 12h + 2h^2 + 1 = 19 + 12h + 2h^2$$

And $f(3) = 19$.

Form the difference quotient: $$\frac{(19 + 12h + 2h^2) - 19}{h} = \frac{12h + 2h^2}{h} = \frac{h(12 + 2h)}{h} = 12 + 2h$$

Take the limit: $$f'(3) = \lim_{h \to 0}(12 + 2h) = \boxed{12}$$

Level 3 Derivative with a Rational Function

Use the limit definition to find $f'(2)$ for $f(x) = \dfrac{4}{x}$.

Thought Process

This is trickier because the function is rational, not polynomial.

  1. Set up $\frac{f(2+h) - f(2)}{h}$
  2. $f(2+h) = \frac{4}{2+h}$ and $f(2) = 2$
  3. To subtract these fractions, find a common denominator
  4. Simplify until you can cancel the $h$
  5. Take the limit

The key algebra step is combining $\frac{4}{2+h} - 2$ into a single fraction.

Show Answer

$$f'(2) = \lim_{h \to 0} \frac{f(2+h) - f(2)}{h} = \lim_{h \to 0} \frac{\frac{4}{2+h} - 2}{h}$$

Find a common denominator for the numerator: $$\frac{4}{2+h} - 2 = \frac{4}{2+h} - \frac{2(2+h)}{2+h} = \frac{4 - 4 - 2h}{2+h} = \frac{-2h}{2+h}$$

Substitute back: $$\frac{\frac{-2h}{2+h}}{h} = \frac{-2h}{h(2+h)} = \frac{-2}{2+h}$$

Take the limit: $$f'(2) = \lim_{h \to 0} \frac{-2}{2+h} = \frac{-2}{2} = \boxed{-1}$$

Level 4 Derivative at a General Point

Use the limit definition to find $f'(a)$ for $f(x) = x^2 - 5x + 2$, where $a$ is any number.

Then verify your answer by computing $f'(4)$ directly from your formula.

Thought Process

Instead of plugging in a specific number for $a$, keep it as a variable throughout the calculation. This gives you a formula for the derivative at any point.

The algebra is similar to Level 2, but you're working with $a$ instead of a number. At the end, you'll have $f'(a)$ as a function of $a$.

To verify: plug $a = 4$ into your formula and check it makes sense.

Show Answer

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

Compute $f(a+h)$: $$f(a+h) = (a+h)^2 - 5(a+h) + 2 = a^2 + 2ah + h^2 - 5a - 5h + 2$$

And $f(a) = a^2 - 5a + 2$.

Form the difference quotient: $$\frac{(a^2 + 2ah + h^2 - 5a - 5h + 2) - (a^2 - 5a + 2)}{h} = \frac{2ah + h^2 - 5h}{h}$$

Factor: $$= \frac{h(2a + h - 5)}{h} = 2a + h - 5$$

Take the limit: $$f'(a) = \lim_{h \to 0}(2a + h - 5) = \boxed{2a - 5}$$

Verification: $f'(4) = 2(4) - 5 = 3$

We can check this makes sense: at $x = 4$, the parabola $f(x) = x^2 - 5x + 2$ should be increasing (since the vertex is at $x = 2.5$), and indeed $f'(4) = 3 > 0$.

Level 5 Derivative of a Square Root Function

Use the limit definition to find $f'(a)$ for $f(x) = \sqrt{x}$, where $a > 0$.

Hint: You'll need to rationalize the numerator.

Thought Process

The direct approach gives: $$\frac{\sqrt{a+h} - \sqrt{a}}{h}$$

This has the form $\frac{0}{0}$ as $h \to 0$. We can't factor out $h$ directly.

The technique: multiply by the conjugate $\frac{\sqrt{a+h} + \sqrt{a}}{\sqrt{a+h} + \sqrt{a}}$ to rationalize the numerator.

This uses the identity: $(A - B)(A + B) = A^2 - B^2$

After rationalizing, the numerator becomes $(a+h) - a = h$, which cancels with the $h$ in the denominator.

Show Answer

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

Multiply by the conjugate: $$= \lim_{h \to 0} \frac{\sqrt{a+h} - \sqrt{a}}{h} \cdot \frac{\sqrt{a+h} + \sqrt{a}}{\sqrt{a+h} + \sqrt{a}}$$

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

$$= \lim_{h \to 0} \frac{h}{h(\sqrt{a+h} + \sqrt{a})}$$

$$= \lim_{h \to 0} \frac{1}{\sqrt{a+h} + \sqrt{a}}$$

Now substitute $h = 0$: $$f'(a) = \frac{1}{\sqrt{a} + \sqrt{a}} = \frac{1}{2\sqrt{a}} = \boxed{\frac{1}{2\sqrt{a}}}$$

This can also be written as $\frac{1}{2}a^{-1/2}$.

Conceptual Questions (CCI-Style)

Level 3 Interpreting the Limit

The limit $\displaystyle\lim_{h \to 0} \frac{(5+h)^4 - 625}{h}$ represents:

(A) The slope of the line through $(5, 625)$ and $(h, h^4)$

(B) The derivative of $f(x) = x^4$ at $x = 5$

(C) The derivative of $f(x) = x^4$ at $x = h$

(D) The average rate of change of $f(x) = x^4$ from $x = 5$ to $x = h$

Thought Process

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

Here we have $\frac{(5+h)^4 - 625}{h}$.

Note that $625 = 5^4$, so this is $\frac{f(5+h) - f(5)}{h}$ where $f(x) = x^4$.

This matches the h-form with $a = 5$.

Show Answer

(B) The derivative of $f(x) = x^4$ at $x = 5$.

Since $625 = 5^4$, the limit has the form: $$\lim_{h \to 0} \frac{f(5+h) - f(5)}{h}$$

which is exactly the definition of $f'(5)$ for $f(x) = x^4$.

Mastery Checklist

Mental Model

The Zoom Lens: Imagine pointing a camera at a curve and zooming in on a single point. The more you zoom, the more the curve looks like a straight line. The derivative is the slope of that line when you've zoomed in "infinitely."

The difference quotient is the slope of the secant line when you're zoomed out. The derivative is what that slope becomes as you zoom in until the two points merge into one.

Connections

Looking back:

Looking ahead:

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Last updated: 2026-01-22