2.6 - Derivative Rules: Constant, Sum, Difference, and Constant Multiple

Three Fundamental Rules

These three rules make taking derivatives way faster. They build on the power rule and let you handle more complicated functions.

Rule 1: Derivative of a Constant

The derivative of any constant is zero.

\frac{d}{dx}[c] = 0

where

c

is any constant.

Why This Makes Sense

Take the function

y = 5

. That’s just a horizontal line.

2.6 - Derivative Rules: Constant, Sum, Difference, and Constant Multiple

What’s the slope of that line? Zero. And since the derivative is the slope,

\frac{dy}{dx} = 0

Examples

Example 1:

f(x) = 8 \Rightarrow f'(x) = 0

Example 2:

g(x) = -15 \Rightarrow g'(x) = 0

Example 3:

h(x) = 742 \Rightarrow h'(x) = 0

Common Mistake: Special Constants

Watch out for these:

\frac{d}{dx}[\pi] = 0 \qquad \frac{d}{dx}[e] = 0 \qquad \frac{d}{dx}[e^2] = 0

Even though

\pi

and

e

are special, they’re still just numbers. Their derivatives are zero. Remember that

\pi \approx 3.14159\ldots

and

e \approx 2.71828\ldots

are constants, not variables.

Rule 2: Constant Multiple Rule

The derivative of a constant times a function is just the constant times the derivative of the function.

\frac{d}{dx}[c \cdot u] = c \cdot \frac{d}{dx}[u]

In words: pull the constant out front, then take the derivative.

Example 4: Basic Constant Multiple

Find the derivative of

y = 7x^3

\frac{dy}{dx} = 7 \cdot \frac{d}{dx}[x^3] = 7 \cdot 3x^2 = 21x^2

The 7 just stays in front while you power-rule the

x^3

Example 5: Negative Constant

Differentiate

f(x) = -4x^5

f'(x) = -4 \cdot 5x^4 = -20x^4

Example 6: Fraction as Constant Multiple

Find the derivative of

y = \dfrac{x^4}{5}

Step 1: Rewrite it as a constant multiple.

y = \frac{1}{5}x^4

Step 2: Apply the rule.

\frac{dy}{dx} = \frac{1}{5} \cdot 4x^3 = \frac{4x^3}{5}

Key idea: dividing by a constant is the same thing as multiplying by its reciprocal.

Rule 3: Sum and Difference Rule

The derivative of a sum or difference is just the sum or difference of the derivatives.

\frac{d}{dx}[u + v] = \frac{d}{dx}[u] + \frac{d}{dx}[v]

\frac{d}{dx}[u - v] = \frac{d}{dx}[u] - \frac{d}{dx}[v]

In words: take the derivative of each term separately.

Example 7: Polynomial with Multiple Terms

Find the derivative of

y = 3x^4 - 7x^2 + 2x - 9

Go term by term:

\frac{dy}{dx} = \frac{d}{dx}[3x^4] - \frac{d}{dx}[7x^2] + \frac{d}{dx}[2x] - \frac{d}{dx}[9]

= 12x^3 - 14x + 2 - 0 = 12x^3 - 14x + 2

Breakdown:

$3x^4 \to 12x^3$ (power rule with a constant)

$-7x^2 \to -14x$ (power rule with a constant)

$2x \to 2$ (power rule: $x^1 \to 1 \cdot x^0 = 1$ )

$-9 \to 0$ (constant rule)

Example 8: Mixed Terms with Fractions

Differentiate

f(x) = 6x - \dfrac{4}{x} + 11

Step 1: Rewrite with negative exponents.

f(x) = 6x - 4x^{-1} + 11

Step 2: Differentiate term by term.

f'(x) = 6 - 4(-1)x^{-2} + 0 = 6 + 4x^{-2}

Step 3: Rewrite with positive exponents.

f'(x) = 6 + \frac{4}{x^2}

Example 9: Radicals and Constants

Find

\frac{dy}{dx}

for

y = 5\sqrt{x} - \dfrac{1}{4}x^7 + 3\pi^2

Step 1: Rewrite.

y = 5x^{1/2} - \frac{1}{4}x^7 + 3\pi^2

Step 2: Differentiate.

\frac{dy}{dx} = 5 \cdot \frac{1}{2}x^{-1/2} - \frac{1}{4} \cdot 7x^6 + 0 = \frac{5}{2}x^{-1/2} - \frac{7}{4}x^6

Step 3: Rewrite.

\frac{dy}{dx} = \frac{5}{2\sqrt{x}} - \frac{7x^6}{4}

Note that

3\pi^2

is a constant (roughly

3(9.87) \approx 29.6

), so its derivative is zero.

Application: Horizontal Tangent Lines

A horizontal tangent line happens wherever the slope equals zero.

To find them:

Find the derivative (the slope function)

Set the derivative equal to zero

Solve for $x$

Example 10: Finding Horizontal Tangents

Find where

f(x) = 2x^2 + 8x - 5

has a horizontal tangent line.

Step 1: Find the derivative.

f'(x) = 4x + 8

Step 2: Set it equal to zero.

4x + 8 = 0

Step 3: Solve.

4x = -8 \quad \Rightarrow \quad x = -2

So there’s a horizontal tangent at

x = -2

, which is the minimum of the parabola.

Application: Normal Lines

A normal line is perpendicular to the tangent line at a given point.

The key concept: perpendicular lines have slopes that are negative reciprocals. If the tangent line has slope

m

, the normal line has slope

-\frac{1}{m}

Example 11: Equation of a Normal Line

Find the equation of the normal line to

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

x = 2

Step 1: Find the point.

f(2) = 8 - 12 + 4 - 5 = -5 \quad \Rightarrow \quad (2, -5)

Step 2: Find the derivative.

f'(x) = 3x^2 - 6x + 2

Step 3: Find the tangent slope.

f'(2) = 3(4) - 6(2) + 2 = 12 - 12 + 2 = 2 \quad \Rightarrow \quad m_{\text{tangent}} = 2

Step 4: Find the normal slope.

m_{\text{normal}} = -\frac{1}{m_{\text{tangent}}} = -\frac{1}{2}

Step 5: Write it in point-slope form.

y + 5 = -\frac{1}{2}(x - 2)

Differentiability of Piecewise Functions

A piecewise function is differentiable at a boundary point if two things are true:

The function is continuous there (the pieces actually connect)

The derivatives are equal there (no sharp corner)

Example 12: Testing Differentiability

Is this differentiable at

x = -2

g(x) = \begin{cases} 3x^2 + 4x + 1 & \text{if } x \leq -2 \\ -5x - 7 & \text{if } x > -2 \end{cases}

Test 1: Continuity. Check whether the pieces meet at

x = -2

Left piece:

3(-2)^2 + 4(-2) + 1 = 12 - 8 + 1 = 5

Right piece:

-5(-2) - 7 = 10 - 7 = 3

Since

5 \neq 3

, it’s not continuous at

x = -2

, so it’s not differentiable there either. You don’t even need the second test.

Example 13: Not Continuous, So Not Differentiable

Is this differentiable at

x = 1

h(x) = \begin{cases} 2x^2 - x + 3 & \text{if } x < 1 \\ x + 4 & \text{if } x \geq 1 \end{cases}

Test 1: Continuity.

Left piece:

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

Right piece:

1 + 4 = 5

Since

4 \neq 5

, not continuous, so not differentiable.

Example 14: Continuous but Not Differentiable

Is this differentiable at

x = 3

f(x) = \begin{cases} x^2 + 2x & \text{if } x \leq 3 \\ -x + 18 & \text{if } x > 3 \end{cases}

Test 1: Continuity.

Left piece:

3^2 + 2(3) = 9 + 6 = 15

Right piece:

-3 + 18 = 15

Since

15 = 15

, the function is continuous. Now test the derivatives.

Test 2: Equal derivatives.

Left piece:

\frac{d}{dx}[x^2 + 2x] = 2x + 2

, and at

x = 3

that’s

2(3) + 2 = 8

Right piece:

\frac{d}{dx}[-x + 18] = -1

, and at

x = 3

that’s

-1

Since

8 \neq -1

, the derivatives don’t match. So the function is continuous but not differentiable at

x = 3

. There’s a sharp corner.

What would make it differentiable? If the right piece were

8x - 9

instead, it would still give

8(3) - 9 = 15

x = 3

(continuous), and its slope would be

8

, matching the left piece. Then both tests pass, and the function would be differentiable at

x = 3

Finding Unknown Constants

Sometimes you have to find the constants that make a piecewise function differentiable.

Example 15: Finding Two Constants

Find

a

and

b

that make this differentiable everywhere:

f(x) = \begin{cases} x^2 + ax - 2 & \text{if } x \leq 1 \\ x + b & \text{if } x > 1 \end{cases}

Condition 1: Continuity at $x = 1$ . Set the pieces equal.

1 + a - 2 = 1 + b \quad \Rightarrow \quad a - 1 = 1 + b \quad \Rightarrow \quad a = b + 2

Condition 2: Equal derivatives at $x = 1$ .

Left derivative:

2x + a

, which at

x = 1

gives

2 + a

Right derivative:

1

2 + a = 1 \quad \Rightarrow \quad a = -1

Solve: Since

a = b + 2

, we get

b = a - 2 = -1 - 2 = -3

Answer:

a = -1

and

b = -3

Practice Problems

Problem 1. Find the derivative:

y = 12x^5 - 3x^2 + 7

$\frac{dy}{dx} = 60x^4 - 6x$

Problem 2. Where does

f(x) = x^2 - 6x + 2

have a horizontal tangent?

$f'(x) = 2x - 6 = 0$ , so $x = 3$

Problem 3. Find the normal line to

g(x) = x^2

x = 3

Point: $(3, 9)$ . $g'(3) = 6$ , so the normal slope is $-\frac{1}{6}$ . Equation: $y - 9 = -\frac{1}{6}(x - 3)$

Key Takeaways

Derivative of a constant is zero: $\frac{d}{dx}[c] = 0$

Constant multiple rule: $\frac{d}{dx}[cu] = c\frac{d}{dx}[u]$

Sum/difference rule: differentiate each term separately

Horizontal tangent happens where $f'(x) = 0$

Normal line is perpendicular to the tangent, so the slopes are negative reciprocals

Differentiability requires both continuity (pieces connect) and equal derivatives (no corner)

Not continuous automatically means not differentiable

Continuous does not guarantee differentiable, since there could be a corner

To find unknowns, set up two equations: continuity and equal derivatives

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