Day 6: Derivative Definition & Basic Rules

The Definition of the Derivative

Everything in Units 2-5 builds on one idea: the derivative is the instantaneous rate of change, obtained by shrinking a secant line down into a tangent line. Once you can compute it from the limit definition, every shortcut rule you learn later is just a proven consequence of this one limit.

Core Theorem

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

Tangent line at

x=a

:

y = f(a) + f'(a)(x - a)

.
Normal line at

x=a

(perpendicular to the tangent):

y = f(a) - \frac{1}{f'(a)}(x - a)

.

Step-by-Step SOP

1
Write the Difference Quotient
Start from $f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$ and substitute the actual function.
2
Algebraically Simplify First
Expand, combine fractions, or rationalize so that the $h$ in the denominator cancels before you take the limit.
3
Take the Limit as h → 0
Only after cancellation should you actually let $h \to 0$ — plugging in $h=0$ too early creates a $\frac{0}{0}$ dead end.

Practice Exercises

Example 01Medium

Use the limit definition of the derivative to find

f'(x)

for

f(x) = x^2

.

NEED A HINT?

Expand

f(x+h) = (x+h)^2

first, then simplify before taking the limit.

SHOW DETAILED EXPLANATION

Step 1: Set Up the Difference Quotient

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

.

Step 2: Expand the Numerator

(x+h)^2 - x^2 = x^2 + 2xh + h^2 - x^2 = 2xh + h^2

.

Step 3: Factor Out h and Cancel

\frac{2xh + h^2}{h} = \frac{h(2x+h)}{h} = 2x + h

(valid since

h \neq 0

inside the limit).

Step 4: Take the Limit

f'(x) = \lim_{h \to 0} (2x + h) = 2x

.

Example 02Easy

Find the equation of the tangent line to

y = x^2

at the point

(2, 4)

using the limit definition.

NEED A HINT?

You already found

f'(x) = 2x

in the previous example — just evaluate it at

x=2

and plug into the point-slope form.

SHOW DETAILED EXPLANATION

Step 1: Find the Slope at x = 2

f'(2) = 2(2) = 4

.

Step 2: Apply Point-Slope Form

y - 4 = 4(x - 2)

.

Step 3: Simplify

y = 4x - 4

.

Example 03Medium

Find both the tangent line and the normal line to

y = \frac{3}{x}

at the point

(3, 1)

.

NEED A HINT?

First find

f'(x)

using the limit definition (or rewrite

\frac{3}{x} = 3x^{-1}

), then build both lines from the point-slope formulas.

SHOW DETAILED EXPLANATION

Step 1: Find f'(x)

f'(x) = \lim_{h \to 0} \frac{\frac{3}{x+h} - \frac{3}{x}}{h} = \lim_{h \to 0} \frac{3x - 3(x+h)}{hx(x+h)} = \lim_{h \to 0} \frac{-3h}{hx(x+h)} = \lim_{h \to 0} \frac{-3}{x(x+h)} = -\frac{3}{x^2}

.

Step 2: Evaluate the Slope at x = 3

f'(3) = -\frac{3}{3^2} = -\frac{3}{9} = -\frac{1}{3}

.

Step 3: Tangent Line

y - 1 = -\frac{1}{3}(x - 3)

.

Step 4: Normal Line

The normal slope is the negative reciprocal of

-\frac{1}{3}

, which is

3

:

y - 1 = 3(x - 3)

.

Example 04Medium

The table below gives selected values of a differentiable function

f

. Estimate

f'(3)

.

$x$	1	2	3	4	5
$f(x)$	2.1	3.4	4.0	5.2	7.5

NEED A HINT?

When estimating a derivative from a table, straddle the target x-value using the closest points on both sides and compute the average rate of change between them.

SHOW DETAILED EXPLANATION

Step 1: Pick the Closest Points on Both Sides

To estimate

f'(3)

, straddle

x=3

using

x=2

and

x=4

rather than a one-sided difference.

Step 2: Compute the Average Rate of Change

f'(3) \approx \frac{f(4) - f(2)}{4 - 2} = \frac{5.2 - 3.4}{2} = \frac{1.8}{2} = 0.9

.

Step 3: Interpret

This symmetric difference quotient is a better estimate than a one-sided one, since it uses information from both directions around

x=3

.

Common Pitfalls

⚠
Don't Skip to the Shortcut Rules YetUntil you've mastered the limit definition, resist reaching for the power rule — AP FRQs sometimes explicitly require 'using the definition of the derivative,' and a shortcut-rule answer earns zero credit there.
⚠
The Normal Line Uses the Negative ReciprocalA common slip is using $f'(a)$ again for the normal line. The normal line's slope is $-\frac{1}{f'(a)}$ , not $f'(a)$ .

Differentiability vs. Continuity

Being continuous means a curve has no breaks. Being differentiable is a stricter promise: the curve also has no sharp corners, no cusps, and no vertical tangent lines. Every differentiable function is continuous, but plenty of continuous functions fail to be differentiable.

Core Theorem

If

f(x)

is differentiable at

x=a

, then

f(x)

is continuous at

x=a

.
A function fails to be differentiable at

x=a

in exactly three ways:
(1)

f

is not continuous at

a

.
(2)

f

is continuous at

a

but

f'_+(a) \neq f'_-(a)

(a corner/kink).
(3)

f

is continuous at

a

but

\lim_{x \to a} |f'(x)| = \infty

(a cusp or vertical tangent).

Step-by-Step SOP

1
Check Continuity First
If a function isn't even continuous at a point, it automatically can't be differentiable there — no need to check derivatives.
2
Compute Both One-Sided Derivatives
Find $f'_-(a)$ and $f'_+(a)$ separately using the appropriate piece of the function on each side.
3
Compare
If the one-sided derivatives are equal and finite, $f$ is differentiable at $a$ . If they disagree, it's a corner; if either blows up to $\pm\infty$ , it's a cusp or vertical tangent.

Practice Exercises

Example 01Easy

Show that

f(x) = |x|

is continuous at

x=0

but not differentiable there.

NEED A HINT?

Compute the left-hand and right-hand derivatives separately using the piecewise definition of

|x|

.

SHOW DETAILED EXPLANATION

Step 1: Continuity is Clear

\lim_{x \to 0} |x| = 0 = f(0)

, so

f

is continuous at

x=0

.

Step 2: Right-Hand Derivative

For

x > 0

,

f(x) = x

, so

f'_+(0) = 1

.

Step 3: Left-Hand Derivative

For

x < 0

,

f(x) = -x

, so

f'_-(0) = -1

.

Step 4: Conclusion

Since

f'_+(0) = 1 \neq -1 = f'_-(0)

, the derivative does not exist at

x=0

— this is a corner.

Example 02Medium

Show that

f(x) = x^{1/3}

is continuous at

x=0

but not differentiable there.

NEED A HINT?

Use the limit definition and watch what happens to the denominator as

h \to 0

.

SHOW DETAILED EXPLANATION

Step 1: Continuity is Clear

x^{1/3}

is defined and continuous for all real

x

, including

x=0

, where

f(0)=0

.

Step 2: Set Up the Difference Quotient

f'(0) = \lim_{h \to 0} \frac{h^{1/3} - 0}{h} = \lim_{h \to 0} \frac{1}{h^{2/3}}

.

Step 3: Evaluate the Limit

As

h \to 0

,

h^{2/3} \to 0^+

, so

\frac{1}{h^{2/3}} \to \infty

.

Step 4: Conclusion

Since the slope blows up to infinity,

f

has a vertical tangent at

x=0

and is not differentiable there — even though the graph has no corner.

Example 03Hard

Let

f(x) = \begin{cases} x^2 & x \le 1 \\ ax + b & x > 1 \end{cases}

. Find the values of

a

and

b

that make

f

differentiable at

x=1

.

NEED A HINT?

Differentiability requires two separate conditions to both hold: continuity at

x=1

, AND matching one-sided derivatives at

x=1

.

SHOW DETAILED EXPLANATION

Step 1: Require Continuity at x = 1

\lim_{x \to 1^-} x^2 = 1

and

\lim_{x \to 1^+} (ax+b) = a+b

. For continuity:

a + b = 1

.

Step 2: Require Matching Derivatives

For

x \le 1

,

f'(x) = 2x

, so

f'_-(1) = 2

. For

x > 1

,

f'(x) = a

, so

f'_+(1) = a

. For differentiability:

a = 2

.

Step 3: Solve the System

Substituting

a=2

into

a+b=1

gives

b = -1

.

Step 4: Final Answer

a = 2

,

b = -1

.

Common Pitfalls

⚠
Continuity Does Not Imply DifferentiabilityThis is the single most tested conceptual fact in this unit. $f(x) = |x|$ is a continuous function that fails to be differentiable — memorize it as your go-to counterexample.
⚠
Matching Value Isn't Enough for Piecewise FunctionsFor a piecewise function to be differentiable at the seam, you need BOTH pieces to agree in value (continuity) AND agree in slope (matching derivatives). Forgetting the slope condition is the most common point loss on this problem type.

Power, Constant & Sum/Difference Rules

Instead of running the limit definition every time, these rules let you differentiate term-by-term almost by inspection. They're proven from the limit definition once, and then used forever after.

Core Theorem

\frac{d}{dx}(c) = 0

\frac{d}{dx}(x^n) = nx^{n-1}

\frac{d}{dx}(cf(x)) = cf'(x)

\frac{d}{dx}(f \pm g) = f' \pm g'

Step-by-Step SOP

1
Rewrite Radicals and Fractions as Powers
Convert every term to the form $x^n$ (including negative and fractional $n$ ) before applying any rule.
2
Differentiate Term by Term
Apply the power rule to each term individually; constants disappear entirely.
3
Match the Question Being Asked
If asked for a horizontal tangent, set $y'=0$ . If asked for a tangent/normal line through a specific point not necessarily on the curve, treat the point of tangency as an unknown and solve for it.

Practice Exercises

Example 01Easy

Differentiate

f(x) = 3x^4 - 5x^2 + 7x - 2

.

NEED A HINT?

Apply the power rule to each term separately, remembering that the derivative of a constant is 0.

SHOW DETAILED EXPLANATION

Step 1: Differentiate Each Term

\frac{d}{dx}(3x^4) = 12x^3

,

\frac{d}{dx}(-5x^2) = -10x

,

\frac{d}{dx}(7x) = 7

,

\frac{d}{dx}(-2) = 0

.

Step 2: Combine

f'(x) = 12x^3 - 10x + 7

.

Example 02Medium

Differentiate

g(x) = 4\sqrt{x} + \frac{3}{x^2}

.

NEED A HINT?

Rewrite both terms as powers of

x

first:

\sqrt{x} = x^{1/2}

and

\frac{1}{x^2} = x^{-2}

.

SHOW DETAILED EXPLANATION

Step 1: Rewrite as Powers

g(x) = 4x^{1/2} + 3x^{-2}

.

Step 2: Apply the Power Rule

\frac{d}{dx}(4x^{1/2}) = 4 \cdot \frac{1}{2}x^{-1/2} = 2x^{-1/2}

, and

\frac{d}{dx}(3x^{-2}) = 3 \cdot (-2)x^{-3} = -6x^{-3}

.

Step 3: Combine and Rewrite

g'(x) = 2x^{-1/2} - 6x^{-3} = \frac{2}{\sqrt{x}} - \frac{6}{x^3}

.

Example 03Medium

Find all points where the tangent line to

y = x^4 - 6x^2 + 4

is horizontal.

NEED A HINT?

A horizontal tangent line means the slope is 0 — set

y'

equal to 0 and solve.

SHOW DETAILED EXPLANATION

Step 1: Differentiate

y' = 4x^3 - 12x

.

Step 2: Set the Derivative to Zero

4x^3 - 12x = 0 \implies 4x(x^2 - 3) = 0 \implies x = 0, \pm\sqrt{3}

.

Step 3: Find the Corresponding y-Values

At

x=0

:

y = 4

. At

x = \pm\sqrt{3}

:

y = 9 - 18 + 4 = -5

.

Step 4: State the Points

Horizontal tangent lines occur at

(0, 4)

,

(\sqrt{3}, -5)

, and

(-\sqrt{3}, -5)

.

Example 04Hard

Find all normal lines to the curve

y = x^3 + 2x^2

that pass through the point

(-2, 0)

.

NEED A HINT?

Let

(a, a^3+2a^2)

be the point of tangency, write the normal line there, then force it to pass through

(-2,0)

and solve for

a

.

SHOW DETAILED EXPLANATION

Step 1: Set Up the General Point and Slope

Let the point of tangency be

(a, a^3+2a^2)

. Since

y' = 3x^2+4x

, the tangent slope there is

3a^2+4a

, so the normal slope is

-\frac{1}{3a^2+4a}

.

Step 2: Write the Normal Line

y - (a^3+2a^2) = -\frac{1}{3a^2+4a}(x - a)

.

Step 3: Force the Line Through (-2, 0)

Substitute

x=-2, y=0

:

-(a^3+2a^2) = -\frac{1}{3a^2+4a}(-2-a)

, which simplifies to

(a^3+2a^2)(3a^2+4a) = -2-a

.

Step 4: Solve and Check a = -1

Testing

a=-1

: left side

= (-1+2)(3-4) = (1)(-1) = -1

, and right side

= -2-(-1) = -1

. It checks out, so

a=-1

works, giving the point of tangency

(-1, 1)

and normal slope

-\frac{1}{3-4} = 1

: the normal line is

y = x+2

.

Common Pitfalls

⚠
Rewrite Before You DifferentiateThe power rule only applies directly to $x^n$ . Radicals and fractions like $\sqrt{x}$ or $\frac{1}{x^2}$ must first be rewritten as $x^{1/2}$ or $x^{-2}$ before you can differentiate them.
⚠
Don't Forget Negative Exponents Flip Sign Changes TwiceWhen the original exponent is negative, both the coefficient multiplication AND the exponent subtraction involve negative numbers — track signs carefully, e.g. $\frac{d}{dx}(x^{-2}) = -2x^{-3}$ , not $2x^{-3}$ .

The Product & Quotient Rules

You cannot differentiate a product or quotient by simply multiplying or dividing the individual derivatives — these two rules tell you exactly how the pieces combine instead.

Core Theorem

Product Rule:

\frac{d}{dx}(fg) = f'g + fg'

.
Quotient Rule:

\frac{d}{dx}\left(\frac{f}{g}\right) = \frac{f'g - fg'}{g^2}

.

Watch the TikTok ExplanationProduct & Quotient Rules→

Step-by-Step SOP

1
Label u and v (or f and g) Clearly
Before differentiating, explicitly identify which factor is which and find their individual derivatives first.
2
Plug into the Formula Exactly
Write out $f'g+fg'$ or $\frac{f'g-fg'}{g^2}$ symbolically first, then substitute — this avoids sign errors under time pressure.
3
Simplify Only If Asked
On the AP exam you often don't need to expand or simplify the final expression — an unsimplified but correct derivative earns full credit.

Practice Exercises

Example 01Easy

Differentiate

f(x) = (x^2+1)(x^3-2x)

using the product rule.

NEED A HINT?

Let

u = x^2+1

and

v = x^3-2x

, then apply

u'v + uv'

.

SHOW DETAILED EXPLANATION

Step 1: Identify u and v

u = x^2+1 \implies u' = 2x

.

v = x^3-2x \implies v' = 3x^2-2

.

Step 2: Apply the Product Rule

f'(x) = u'v + uv' = 2x(x^3-2x) + (x^2+1)(3x^2-2)

.

Step 3: Expand and Simplify

= (2x^4 - 4x^2) + (3x^4 - 2x^2 + 3x^2 - 2) = 5x^4 - 3x^2 - 2

.

Example 02Medium

Differentiate

g(x) = \frac{x^2 - 1}{x^2 + 1}

using the quotient rule.

NEED A HINT?

Let

u = x^2-1

(top) and

v = x^2+1

(bottom), then apply

\frac{u'v-uv'}{v^2}

.

SHOW DETAILED EXPLANATION

Step 1: Identify u and v

u = x^2-1 \implies u'=2x

.

v = x^2+1 \implies v'=2x

.

Step 2: Apply the Quotient Rule

g'(x) = \frac{2x(x^2+1) - (x^2-1)(2x)}{(x^2+1)^2}

.

Step 3: Expand the Numerator

2x(x^2+1) - 2x(x^2-1) = 2x[(x^2+1)-(x^2-1)] = 2x(2) = 4x

.

Step 4: Final Answer

g'(x) = \frac{4x}{(x^2+1)^2}

.

Example 03Medium

If

y = uv

and

u(2)=3

,

u'(2)=-4

,

v(2)=1

,

v'(2)=2

, find

y'(2)

.

NEED A HINT?

You don't need formulas for

u

or

v

themselves — just plug the given values directly into the product rule.

SHOW DETAILED EXPLANATION

Step 1: Write the Product Rule

y' = u'v + uv'

.

Step 2: Substitute the Given Values at x = 2

y'(2) = u'(2)v(2) + u(2)v'(2) = (-4)(1) + (3)(2)

.

Step 3: Compute

y'(2) = -4 + 6 = 2

.

Example 04Hard

If

f(x) = \sqrt{x}\,g(x)

, and

g(4) = 2

,

g'(4) = 3

, find

f'(4)

.

NEED A HINT?

Treat

\sqrt{x}

as one factor and

g(x)

as the other, differentiate

\sqrt{x}

with the power rule, then apply the product rule.

SHOW DETAILED EXPLANATION

Step 1: Set Up the Product Rule

f(x) = x^{1/2} g(x) \implies f'(x) = \frac{1}{2}x^{-1/2}g(x) + x^{1/2}g'(x)

.

Step 2: Substitute x = 4

f'(4) = \frac{1}{2}(4)^{-1/2}g(4) + (4)^{1/2}g'(4) = \frac{1}{2}\cdot\frac{1}{2}\cdot 2 + 2 \cdot 3

.

Step 3: Compute

f'(4) = \frac{1}{2} + 6 = \frac{13}{2}

.

Common Pitfalls

⚠
The Product of Derivatives Is NOT the Derivative of the Product $\frac{d}{dx}(fg) \neq f' \cdot g'$ . This is the single most common error in this unit — always use $f'g+fg'$ .
⚠
Quotient Rule Order MattersThe numerator is $f'g - fg'$ , not $fg' - f'g$ — reversing the order flips the sign of your answer. Remember the mnemonic 'low-D-high minus high-D-low.'

Write the Difference Quotient

Algebraically Simplify First

Take the Limit as h → 0

Practice Exercises

Step 1: Set Up the Difference Quotient

Step 2: Expand the Numerator

Step 3: Factor Out h and Cancel

Step 4: Take the Limit

Step 1: Find the Slope at x = 2

Step 2: Apply Point-Slope Form

Step 3: Simplify

Step 1: Find f'(x)

Step 2: Evaluate the Slope at x = 3

Step 3: Tangent Line

Step 4: Normal Line

Step 1: Pick the Closest Points on Both Sides

Step 2: Compute the Average Rate of Change

Step 3: Interpret

Check Continuity First

Compute Both One-Sided Derivatives

Compare

Practice Exercises

Step 1: Continuity is Clear

Step 2: Right-Hand Derivative

Step 3: Left-Hand Derivative

Step 4: Conclusion

Step 1: Continuity is Clear

Step 2: Set Up the Difference Quotient

Step 3: Evaluate the Limit

Step 4: Conclusion

Step 1: Require Continuity at x = 1

Step 2: Require Matching Derivatives

Step 3: Solve the System

Step 4: Final Answer

Rewrite Radicals and Fractions as Powers

Differentiate Term by Term

Match the Question Being Asked

Practice Exercises

Step 1: Differentiate Each Term

Step 2: Combine

Step 1: Rewrite as Powers

Step 2: Apply the Power Rule

Step 3: Combine and Rewrite

Step 1: Differentiate

Step 2: Set the Derivative to Zero

Step 3: Find the Corresponding y-Values

Step 4: State the Points

Step 1: Set Up the General Point and Slope

Step 2: Write the Normal Line

Step 3: Force the Line Through (-2, 0)

Step 4: Solve and Check a = -1

Label u and v (or f and g) Clearly

Plug into the Formula Exactly

Simplify Only If Asked

Practice Exercises

Step 1: Identify u and v

Step 2: Apply the Product Rule

Step 3: Expand and Simplify

Step 1: Identify u and v

Step 2: Apply the Quotient Rule

Step 3: Expand the Numerator

Step 4: Final Answer

Step 1: Write the Product Rule

Step 2: Substitute the Given Values at x = 2

Step 3: Compute

Step 1: Set Up the Product Rule

Step 2: Substitute x = 4

Step 3: Compute

Gary Chang