2.3 Power rule

Achievable AP Calculus AB

2. Derivative basics

Our AP Calculus AB course is currently in development and is a work-in-progress.

Power rule

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What you’ll learn:

How to compute derivatives of functions quickly with the power rule
Differentiating constants, linear terms, and sums or differences

Instead of using the limit definition every time (which can get messy for more complicated functions), you can use derivative rules to find derivatives quickly. One of the most useful is the power rule, which applies to functions where a variable is raised to a constant power.

The power rule

When differentiating functions of the form $x^{n}$ , where $n$ is any real number, apply the power rule.

Power rule:

The derivative of $x^{n}$ is:

$\frac{d}{d x} x^{n} = n x^{n - 1}$

To differentiate $x^{n}$ :

Multiply by the exponent $n$ .
Subtract $1$ from the exponent.

For example,

$\frac{d}{d x} (x^{6}) = 6 x^{6 - 1} = 6 x^{5}$

$\frac{d}{d x} (x^{3}) = 3 x^{3 - 1} = 3 x^{2}$

Most functions you’ll work with won’t be a single $x^{n}$ term. Fortunately, a few additional rules make it straightforward to differentiate expressions with constants, sums, and differences.

One useful extension is the constant multiple rule.

Constant multiple rule:

$\frac{d}{d x} [c \cdot f (x)] = c \cdot \frac{d}{d x} [f (x)]$

where $c$ is a real number.

This rule says you can “pull out” a constant and multiply it by the derivative at the end.

For example,

$\frac{d}{d x} (- 5 x^{2})$

(spoiler)

$= - 5 \cdot \frac{d}{d x} (x^{2})$

$= - 5 \cdot (2 x^{2 - 1})$

$= - 10 x$

$\frac{d}{d x} [(2 x)^{3}]$

(spoiler)

$= \frac{d}{d x} (8 x^{3})$

$= 8 \cdot \frac{d}{d x} (x^{3})$

$= 8 \cdot 3 x^{3 - 1}$

$= 24 x^{2}$

Another rule is the sum/difference rule. When a function is made up of terms added or subtracted, differentiate each term separately.

Sum/difference rule:

If $f (x) = g (x) \pm h (x),$ then $f^{'} (x) = g^{'} (x) \pm h^{'} (x)$

Example

The derivative of the polynomial $f (x) = x^{4} - 3 x^{2}$ is:

(spoiler)

$f^{'} (x) = 4 x^{4 - 1} - 3 \cdot (2 x^{2 - 1}) = 4 x^{3} - 6 x$

The power rule also works for terms that are linear, constant, or have negative or fractional exponents.

Linear terms

Differentiate $f (x) = 4 x$ , noting that $x$ is actually $x^{1}$ .

(spoiler)

$\frac{d}{d x} (4 x^{1})$

$= 1 \cdot (4 x^{1 - 1})$

$= 4 x^{0}$

$= 4$

So the derivative of a linear term $a x$ is just its coefficient $a$ . This matches the geometry: the graph of $y = 4 x$ is a straight line, and its slope is $4$ everywhere.

When f(x) = a constant*

The derivative of a constant is always $0$ . You can connect this to the power rule by writing a constant as something times $x^{0}$ , since $x^{0} = 1$ .

Differentiate $f (x) = 2$ .

(spoiler)

$\frac{d}{d x} (2)$

$= \frac{d}{d x} (2 x^{0})$

$= 0 x^{0 - 1}$

$= 0$

This also fits the graph: a constant function is a horizontal line, and horizontal lines have slope $0$ .

What is the derivative of $g (x) = π^{2}$ ?

(spoiler)

Although $π$ looks like a variable, it’s a constant ( $\approx 3.14$ ). Therefore,

$\frac{d}{d x} (π^{2}) = 0$

Negative exponents

Use the power rule to differentiate $f (x) = x^{- 4}$ .

(spoiler)

$\frac{d}{d x} (x^{- 4})$

$= - 4 x^{- 4 - 1}$

$= - 4 x^{- 5}$

$= - \frac{4}{x ^{5}}$

AP tip:

Try to spot functions that are quotients with a number on top and an expression with a power on the bottom. Those can be rewritten in the form $x^{n}$ (where $n$ is negative). For example, $\frac{2}{x ^{3}}$ becomes $2 x^{- 3}$ before you apply the power rule.

Fractional exponents

Use the power rule to differentiate $f (x) = x^{1/2}$ .

(spoiler)

$\frac{d}{d x} (x^{1/2})$

$= \frac{1}{2} x^{(1/2) - 1}$

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

$= \frac{1}{2 x ^{1/2}}$

$= \frac{1}{2 x}$

AP tip:

Whenever you see a radical function, rewrite it using fractional exponents so you can apply the power rule. For example, $3 x = x^{1/3}$ and $4 x^{3} = x^{3/4}$ .

Try to simplify expressions when you can. For example, the expression

$\frac{x ^{3} + 3 x}{2 x}$

can be split into

$\frac{x ^{3}}{3 x} + \frac{3 x}{2 x}$

and simplified into

$\frac{1}{3} x^{2} + \frac{3}{2}$

Later on, you’ll learn how to differentiate quotients directly, but sometimes rewriting and simplifying lets you use the power rule instead.

::: sidenote When the power rule can’t be used Keep in mind the power rule only applies to functions where the exponent is a constant, not a variable. For example:

$\frac{d}{d x} (2^{x}) \neq = x \cdot 2^{x - 1}$

because $2^{x}$ is an exponential function, not a power function. :::

The correct derivative will be covered in section 3.5 (Special derivatives), where you’ll learn how to handle derivatives of exponentials and other functions.

AP tip:

Remember that symbols such as $π$ and $e$ are constants ( $π \approx 3.14$ and $e \approx 2.72$ ), not variables. So the power rule can be applied if those are in the exponent. For example, if $f (x) = x^{e}$ , then $e$ is a constant and the derivative is $f^{'} (x) = e \cdot x^{e - 1}$ .

Challenge problems

Find the derivative of $f (x) = \frac{1}{3 x ^{2}} + 2 x - 1$

Solution

(spoiler)

Answer: $f^{'} (x) = - \frac{2}{3 x ^{5/3}} + \frac{1}{x ^{1/2}}$

Rewrite the radical terms using fractional exponents.

$\frac{1}{3 x ^{2}} = \frac{1}{x ^{2/3}} = x^{- 2/3}$

and

$2 x = 2 x^{1/2}$

Now the function is

$f (x) = x^{- 2/3} + 2 x^{1/2} - 1$

Apply the power rule to each term.

$\frac{d}{d x} f (x) = \frac{d}{d x} (x^{- 2/3}) + 2 \frac{d}{d x} (x^{1/2}) - \frac{d}{d x} (1) = - \frac{2}{3} x^{- 5/3} + 2 \cdot \frac{1}{2} x^{- 1/2} - 0 = - \frac{2}{3 x ^{5/3}} + \frac{1}{x ^{1/2}}$

Find the value(s) of $k$ such that $y = 3 x + k$ is a tangent line to $f (x) = x^{3}$ .

Hint: At the point of tangency, a function and its tangent line share:

The $y$ -value

The slope value

Solution

(spoiler)

Answer: $k = 2 or - 2$

It helps to sketch these two functions to see what the question is asking.

Using the power rule, the derivative is $f^{'} (x) = 3 x^{2}$ .

At a point of tangency $(a, f (a))$ , the function has the same slope as the tangent line.

The line $y = 3 x + k$ has slope $3$ . The function has slope $3$ when $f^{'} (a) = 3$ .

$3 a^{2} = 3$

$a^{2} = 1$

$a = \pm 1$

So the points of tangency are $(1, f (1)) = (1, 1)$ and $(- 1, f (- 1)) = (- 1, - 1)$ .

At each tangency point, the line and the function also share the same $y$ -value, so substitute $x = a$ into $y = 3 x + k$ and match it to $f (a)$ .

For $(1, 1)$ :

$3 (1) + k = (1)^{3}$

$k = - 2$

For $(- 1, - 1)$ :

$3 (- 1) + k = (- 1)^{3}$

$k = 2$

Find the derivative of $y = (x + 1) (x^{3} - 2)$

Solution

(spoiler)

Answer: $y^{'} = 4 x^{3} + 3 x^{2} - 2$

First, expand so the function is written as a sum of power terms.

$y = x (x^{3}) + x (- 2) + 1 (x^{3}) + 1 (- 2)$

$= x^{4} - 2 x + x^{3} - 2$

Next, apply the power rule to each term.

$y^{'} = 4 x^{3} - 2 + 3 x^{2} - 0$

$= 4 x^{3} + 3 x^{2} - 2$

In the next section, you’ll learn how to do this same problem using the product rule for the product of two functions.

Power rule

Derivative of $x^{n}$ is $n x^{n - 1}$
Multiply by exponent, subtract 1 from exponent
Applies for any real exponent $n$

Constant multiple rule

Derivative of $c \cdot f (x)$ is $c \cdot f^{'} (x)$
Constants can be factored out before differentiating

Sum/difference rule

Derivative of $f (x) = g (x) \pm h (x)$ is $g^{'} (x) \pm h^{'} (x)$
Differentiate each term separately

Special cases for the power rule

Linear terms: derivative of $a x$ is $a$
Constant terms: derivative is $0$
- Any constant (e.g., $2$ , $π^{2}$ )
Negative exponents: apply power rule, result may be negative exponent or reciprocal
Fractional exponents: rewrite radicals as $x^{m / n}$ , apply power rule

Simplifying before differentiating

Rewrite quotients and radicals as power functions when possible
Simplify expressions to sums of power terms for easier differentiation

When the power rule does not apply

Power rule only for $x^{n}$ where $n$ is constant
Does not apply to exponential functions like $2^{x}$

Constants in exponents

If exponent is a constant (e.g., $x^{e}$ ), power rule applies
$π$ and $e$ are constants, not variables

AP tips

Rewrite radicals and quotients using exponents for easier differentiation
Recognize constants vs. variables in expressions

Challenge problem strategies

Convert all terms to power form before differentiating
For tangency: match both slope and $y$ -value at the point
Expand products before applying power rule to each term

Power rule

What you’ll learn:

How to compute derivatives of functions quickly with the power rule
Differentiating constants, linear terms, and sums or differences

The power rule

When differentiating functions of the form $x^{n}$ , where $n$ is any real number, apply the power rule.

Power rule:

The derivative of $x^{n}$ is:

$\frac{d}{d x} x^{n} = n x^{n - 1}$

To differentiate $x^{n}$ :

Multiply by the exponent $n$ .
Subtract $1$ from the exponent.

For example,

$\frac{d}{d x} (x^{6}) = 6 x^{6 - 1} = 6 x^{5}$

$\frac{d}{d x} (x^{3}) = 3 x^{3 - 1} = 3 x^{2}$

Most functions you’ll work with won’t be a single $x^{n}$ term. Fortunately, a few additional rules make it straightforward to differentiate expressions with constants, sums, and differences.

One useful extension is the constant multiple rule.

Constant multiple rule:

$\frac{d}{d x} [c \cdot f (x)] = c \cdot \frac{d}{d x} [f (x)]$

where $c$ is a real number.

This rule says you can “pull out” a constant and multiply it by the derivative at the end.

For example,

$\frac{d}{d x} (- 5 x^{2})$

(spoiler)

$= - 5 \cdot \frac{d}{d x} (x^{2})$

$= - 5 \cdot (2 x^{2 - 1})$

$= - 10 x$

$\frac{d}{d x} [(2 x)^{3}]$

(spoiler)

$= \frac{d}{d x} (8 x^{3})$

$= 8 \cdot \frac{d}{d x} (x^{3})$

$= 8 \cdot 3 x^{3 - 1}$

$= 24 x^{2}$

Another rule is the sum/difference rule. When a function is made up of terms added or subtracted, differentiate each term separately.

Sum/difference rule:

If $f (x) = g (x) \pm h (x),$ then $f^{'} (x) = g^{'} (x) \pm h^{'} (x)$

Example

The derivative of the polynomial $f (x) = x^{4} - 3 x^{2}$ is:

(spoiler)

$f^{'} (x) = 4 x^{4 - 1} - 3 \cdot (2 x^{2 - 1}) = 4 x^{3} - 6 x$

The power rule also works for terms that are linear, constant, or have negative or fractional exponents.

Linear terms

Differentiate $f (x) = 4 x$ , noting that $x$ is actually $x^{1}$ .

(spoiler)

$\frac{d}{d x} (4 x^{1})$

$= 1 \cdot (4 x^{1 - 1})$

$= 4 x^{0}$

$= 4$

So the derivative of a linear term $a x$ is just its coefficient $a$ . This matches the geometry: the graph of $y = 4 x$ is a straight line, and its slope is $4$ everywhere.

When f(x) = a constant*

The derivative of a constant is always $0$ . You can connect this to the power rule by writing a constant as something times $x^{0}$ , since $x^{0} = 1$ .

Differentiate $f (x) = 2$ .

(spoiler)

$\frac{d}{d x} (2)$

$= \frac{d}{d x} (2 x^{0})$

$= 0 x^{0 - 1}$

$= 0$

This also fits the graph: a constant function is a horizontal line, and horizontal lines have slope $0$ .

What is the derivative of $g (x) = π^{2}$ ?

(spoiler)

Although $π$ looks like a variable, it’s a constant ( $\approx 3.14$ ). Therefore,

$\frac{d}{d x} (π^{2}) = 0$

Negative exponents

Use the power rule to differentiate $f (x) = x^{- 4}$ .

(spoiler)

$\frac{d}{d x} (x^{- 4})$

$= - 4 x^{- 4 - 1}$

$= - 4 x^{- 5}$

$= - \frac{4}{x ^{5}}$

AP tip:

Fractional exponents

Use the power rule to differentiate $f (x) = x^{1/2}$ .

(spoiler)

$\frac{d}{d x} (x^{1/2})$

$= \frac{1}{2} x^{(1/2) - 1}$

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

$= \frac{1}{2 x ^{1/2}}$

$= \frac{1}{2 x}$

AP tip:

Whenever you see a radical function, rewrite it using fractional exponents so you can apply the power rule. For example, $3 x = x^{1/3}$ and $4 x^{3} = x^{3/4}$ .

Try to simplify expressions when you can. For example, the expression

$\frac{x ^{3} + 3 x}{2 x}$

can be split into

$\frac{x ^{3}}{3 x} + \frac{3 x}{2 x}$

and simplified into

$\frac{1}{3} x^{2} + \frac{3}{2}$

Later on, you’ll learn how to differentiate quotients directly, but sometimes rewriting and simplifying lets you use the power rule instead.

::: sidenote When the power rule can’t be used Keep in mind the power rule only applies to functions where the exponent is a constant, not a variable. For example:

$\frac{d}{d x} (2^{x}) \neq = x \cdot 2^{x - 1}$

because $2^{x}$ is an exponential function, not a power function. :::

The correct derivative will be covered in section 3.5 (Special derivatives), where you’ll learn how to handle derivatives of exponentials and other functions.

AP tip:

Challenge problems

Find the derivative of $f (x) = \frac{1}{3 x ^{2}} + 2 x - 1$

Solution

(spoiler)

Answer: $f^{'} (x) = - \frac{2}{3 x ^{5/3}} + \frac{1}{x ^{1/2}}$

Rewrite the radical terms using fractional exponents.

$\frac{1}{3 x ^{2}} = \frac{1}{x ^{2/3}} = x^{- 2/3}$

and

$2 x = 2 x^{1/2}$

Now the function is

$f (x) = x^{- 2/3} + 2 x^{1/2} - 1$

Apply the power rule to each term.

Find the value(s) of $k$ such that $y = 3 x + k$ is a tangent line to $f (x) = x^{3}$ .

Hint: At the point of tangency, a function and its tangent line share:

The $y$ -value

The slope value

Solution

(spoiler)

Answer: $k = 2 or - 2$

It helps to sketch these two functions to see what the question is asking.

Using the power rule, the derivative is $f^{'} (x) = 3 x^{2}$ .

At a point of tangency $(a, f (a))$ , the function has the same slope as the tangent line.

The line $y = 3 x + k$ has slope $3$ . The function has slope $3$ when $f^{'} (a) = 3$ .

$3 a^{2} = 3$

$a^{2} = 1$

$a = \pm 1$

So the points of tangency are $(1, f (1)) = (1, 1)$ and $(- 1, f (- 1)) = (- 1, - 1)$ .

At each tangency point, the line and the function also share the same $y$ -value, so substitute $x = a$ into $y = 3 x + k$ and match it to $f (a)$ .

For $(1, 1)$ :

$3 (1) + k = (1)^{3}$

$k = - 2$

For $(- 1, - 1)$ :

$3 (- 1) + k = (- 1)^{3}$

$k = 2$

Find the derivative of $y = (x + 1) (x^{3} - 2)$

Solution

(spoiler)

Answer: $y^{'} = 4 x^{3} + 3 x^{2} - 2$

First, expand so the function is written as a sum of power terms.

$y = x (x^{3}) + x (- 2) + 1 (x^{3}) + 1 (- 2)$

$= x^{4} - 2 x + x^{3} - 2$

Next, apply the power rule to each term.

$y^{'} = 4 x^{3} - 2 + 3 x^{2} - 0$

$= 4 x^{3} + 3 x^{2} - 2$

In the next section, you’ll learn how to do this same problem using the product rule for the product of two functions.

Achievable AP Calculus AB

2. Derivative basics

Our AP Calculus AB course is currently in development and is a work-in-progress.

Power rule

7 min read

Font

Discuss

Feedback

What you’ll learn:

How to compute derivatives of functions quickly with the power rule
Differentiating constants, linear terms, and sums or differences

The power rule

When differentiating functions of the form $x^{n}$ , where $n$ is any real number, apply the power rule.

Power rule:

The derivative of $x^{n}$ is:

$\frac{d}{d x} x^{n} = n x^{n - 1}$

To differentiate $x^{n}$ :

Multiply by the exponent $n$ .
Subtract $1$ from the exponent.

For example,

$\frac{d}{d x} (x^{6}) = 6 x^{6 - 1} = 6 x^{5}$

$\frac{d}{d x} (x^{3}) = 3 x^{3 - 1} = 3 x^{2}$

Most functions you’ll work with won’t be a single $x^{n}$ term. Fortunately, a few additional rules make it straightforward to differentiate expressions with constants, sums, and differences.

One useful extension is the constant multiple rule.

Constant multiple rule:

$\frac{d}{d x} [c \cdot f (x)] = c \cdot \frac{d}{d x} [f (x)]$

where $c$ is a real number.

This rule says you can “pull out” a constant and multiply it by the derivative at the end.

For example,

$\frac{d}{d x} (- 5 x^{2})$

(spoiler)

$= - 5 \cdot \frac{d}{d x} (x^{2})$

$= - 5 \cdot (2 x^{2 - 1})$

$= - 10 x$

$\frac{d}{d x} [(2 x)^{3}]$

(spoiler)

$= \frac{d}{d x} (8 x^{3})$

$= 8 \cdot \frac{d}{d x} (x^{3})$

$= 8 \cdot 3 x^{3 - 1}$

$= 24 x^{2}$

Another rule is the sum/difference rule. When a function is made up of terms added or subtracted, differentiate each term separately.

Sum/difference rule:

If $f (x) = g (x) \pm h (x),$ then $f^{'} (x) = g^{'} (x) \pm h^{'} (x)$

Example

The derivative of the polynomial $f (x) = x^{4} - 3 x^{2}$ is:

(spoiler)

$f^{'} (x) = 4 x^{4 - 1} - 3 \cdot (2 x^{2 - 1}) = 4 x^{3} - 6 x$

The power rule also works for terms that are linear, constant, or have negative or fractional exponents.

Linear terms

Differentiate $f (x) = 4 x$ , noting that $x$ is actually $x^{1}$ .

(spoiler)

$\frac{d}{d x} (4 x^{1})$

$= 1 \cdot (4 x^{1 - 1})$

$= 4 x^{0}$

$= 4$

So the derivative of a linear term $a x$ is just its coefficient $a$ . This matches the geometry: the graph of $y = 4 x$ is a straight line, and its slope is $4$ everywhere.

When f(x) = a constant*

The derivative of a constant is always $0$ . You can connect this to the power rule by writing a constant as something times $x^{0}$ , since $x^{0} = 1$ .

Differentiate $f (x) = 2$ .

(spoiler)

$\frac{d}{d x} (2)$

$= \frac{d}{d x} (2 x^{0})$

$= 0 x^{0 - 1}$

$= 0$

This also fits the graph: a constant function is a horizontal line, and horizontal lines have slope $0$ .

What is the derivative of $g (x) = π^{2}$ ?

(spoiler)

Although $π$ looks like a variable, it’s a constant ( $\approx 3.14$ ). Therefore,

$\frac{d}{d x} (π^{2}) = 0$

Negative exponents

Use the power rule to differentiate $f (x) = x^{- 4}$ .

(spoiler)

$\frac{d}{d x} (x^{- 4})$

$= - 4 x^{- 4 - 1}$

$= - 4 x^{- 5}$

$= - \frac{4}{x ^{5}}$

AP tip:

Fractional exponents

Use the power rule to differentiate $f (x) = x^{1/2}$ .

(spoiler)

$\frac{d}{d x} (x^{1/2})$

$= \frac{1}{2} x^{(1/2) - 1}$

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

$= \frac{1}{2 x ^{1/2}}$

$= \frac{1}{2 x}$

AP tip:

Whenever you see a radical function, rewrite it using fractional exponents so you can apply the power rule. For example, $3 x = x^{1/3}$ and $4 x^{3} = x^{3/4}$ .

Try to simplify expressions when you can. For example, the expression

$\frac{x ^{3} + 3 x}{2 x}$

can be split into

$\frac{x ^{3}}{3 x} + \frac{3 x}{2 x}$

and simplified into

$\frac{1}{3} x^{2} + \frac{3}{2}$

Later on, you’ll learn how to differentiate quotients directly, but sometimes rewriting and simplifying lets you use the power rule instead.

::: sidenote When the power rule can’t be used Keep in mind the power rule only applies to functions where the exponent is a constant, not a variable. For example:

$\frac{d}{d x} (2^{x}) \neq = x \cdot 2^{x - 1}$

because $2^{x}$ is an exponential function, not a power function. :::

The correct derivative will be covered in section 3.5 (Special derivatives), where you’ll learn how to handle derivatives of exponentials and other functions.

AP tip:

Challenge problems

Find the derivative of $f (x) = \frac{1}{3 x ^{2}} + 2 x - 1$

Solution

(spoiler)

Answer: $f^{'} (x) = - \frac{2}{3 x ^{5/3}} + \frac{1}{x ^{1/2}}$

Rewrite the radical terms using fractional exponents.

$\frac{1}{3 x ^{2}} = \frac{1}{x ^{2/3}} = x^{- 2/3}$

and

$2 x = 2 x^{1/2}$

Now the function is

$f (x) = x^{- 2/3} + 2 x^{1/2} - 1$

Apply the power rule to each term.

Find the value(s) of $k$ such that $y = 3 x + k$ is a tangent line to $f (x) = x^{3}$ .

Hint: At the point of tangency, a function and its tangent line share:

The $y$ -value

The slope value

Solution

(spoiler)

Answer: $k = 2 or - 2$

It helps to sketch these two functions to see what the question is asking.

Using the power rule, the derivative is $f^{'} (x) = 3 x^{2}$ .

At a point of tangency $(a, f (a))$ , the function has the same slope as the tangent line.

The line $y = 3 x + k$ has slope $3$ . The function has slope $3$ when $f^{'} (a) = 3$ .

$3 a^{2} = 3$

$a^{2} = 1$

$a = \pm 1$

So the points of tangency are $(1, f (1)) = (1, 1)$ and $(- 1, f (- 1)) = (- 1, - 1)$ .

At each tangency point, the line and the function also share the same $y$ -value, so substitute $x = a$ into $y = 3 x + k$ and match it to $f (a)$ .

For $(1, 1)$ :

$3 (1) + k = (1)^{3}$

$k = - 2$

For $(- 1, - 1)$ :

$3 (- 1) + k = (- 1)^{3}$

$k = 2$

Find the derivative of $y = (x + 1) (x^{3} - 2)$

Solution

(spoiler)

Answer: $y^{'} = 4 x^{3} + 3 x^{2} - 2$

First, expand so the function is written as a sum of power terms.

$y = x (x^{3}) + x (- 2) + 1 (x^{3}) + 1 (- 2)$

$= x^{4} - 2 x + x^{3} - 2$

Next, apply the power rule to each term.

$y^{'} = 4 x^{3} - 2 + 3 x^{2} - 0$

$= 4 x^{3} + 3 x^{2} - 2$

In the next section, you’ll learn how to do this same problem using the product rule for the product of two functions.

Power rule

Derivative of $x^{n}$ is $n x^{n - 1}$
Multiply by exponent, subtract 1 from exponent
Applies for any real exponent $n$

Constant multiple rule

Derivative of $c \cdot f (x)$ is $c \cdot f^{'} (x)$
Constants can be factored out before differentiating

Sum/difference rule

Derivative of $f (x) = g (x) \pm h (x)$ is $g^{'} (x) \pm h^{'} (x)$
Differentiate each term separately

Special cases for the power rule

Linear terms: derivative of $a x$ is $a$
Constant terms: derivative is $0$
- Any constant (e.g., $2$ , $π^{2}$ )
Negative exponents: apply power rule, result may be negative exponent or reciprocal
Fractional exponents: rewrite radicals as $x^{m / n}$ , apply power rule

Simplifying before differentiating

Rewrite quotients and radicals as power functions when possible
Simplify expressions to sums of power terms for easier differentiation

When the power rule does not apply

Power rule only for $x^{n}$ where $n$ is constant
Does not apply to exponential functions like $2^{x}$

Constants in exponents

If exponent is a constant (e.g., $x^{e}$ ), power rule applies
$π$ and $e$ are constants, not variables

AP tips

Rewrite radicals and quotients using exponents for easier differentiation
Recognize constants vs. variables in expressions

Challenge problem strategies

Convert all terms to power form before differentiating
For tangency: match both slope and $y$ -value at the point
Expand products before applying power rule to each term

Power rule

What you’ll learn:

How to compute derivatives of functions quickly with the power rule
Differentiating constants, linear terms, and sums or differences

The power rule

When differentiating functions of the form $x^{n}$ , where $n$ is any real number, apply the power rule.

Power rule:

The derivative of $x^{n}$ is:

$\frac{d}{d x} x^{n} = n x^{n - 1}$

To differentiate $x^{n}$ :

Multiply by the exponent $n$ .
Subtract $1$ from the exponent.

For example,

$\frac{d}{d x} (x^{6}) = 6 x^{6 - 1} = 6 x^{5}$

$\frac{d}{d x} (x^{3}) = 3 x^{3 - 1} = 3 x^{2}$

Most functions you’ll work with won’t be a single $x^{n}$ term. Fortunately, a few additional rules make it straightforward to differentiate expressions with constants, sums, and differences.

One useful extension is the constant multiple rule.

Constant multiple rule:

$\frac{d}{d x} [c \cdot f (x)] = c \cdot \frac{d}{d x} [f (x)]$

where $c$ is a real number.

This rule says you can “pull out” a constant and multiply it by the derivative at the end.

For example,

$\frac{d}{d x} (- 5 x^{2})$

(spoiler)

$= - 5 \cdot \frac{d}{d x} (x^{2})$

$= - 5 \cdot (2 x^{2 - 1})$

$= - 10 x$

$\frac{d}{d x} [(2 x)^{3}]$

(spoiler)

$= \frac{d}{d x} (8 x^{3})$

$= 8 \cdot \frac{d}{d x} (x^{3})$

$= 8 \cdot 3 x^{3 - 1}$

$= 24 x^{2}$

Another rule is the sum/difference rule. When a function is made up of terms added or subtracted, differentiate each term separately.

Sum/difference rule:

If $f (x) = g (x) \pm h (x),$ then $f^{'} (x) = g^{'} (x) \pm h^{'} (x)$

Example

The derivative of the polynomial $f (x) = x^{4} - 3 x^{2}$ is:

(spoiler)

$f^{'} (x) = 4 x^{4 - 1} - 3 \cdot (2 x^{2 - 1}) = 4 x^{3} - 6 x$

The power rule also works for terms that are linear, constant, or have negative or fractional exponents.

Linear terms

Differentiate $f (x) = 4 x$ , noting that $x$ is actually $x^{1}$ .

(spoiler)

$\frac{d}{d x} (4 x^{1})$

$= 1 \cdot (4 x^{1 - 1})$

$= 4 x^{0}$

$= 4$

So the derivative of a linear term $a x$ is just its coefficient $a$ . This matches the geometry: the graph of $y = 4 x$ is a straight line, and its slope is $4$ everywhere.

When f(x) = a constant*

The derivative of a constant is always $0$ . You can connect this to the power rule by writing a constant as something times $x^{0}$ , since $x^{0} = 1$ .

Differentiate $f (x) = 2$ .

(spoiler)

$\frac{d}{d x} (2)$

$= \frac{d}{d x} (2 x^{0})$

$= 0 x^{0 - 1}$

$= 0$

This also fits the graph: a constant function is a horizontal line, and horizontal lines have slope $0$ .

What is the derivative of $g (x) = π^{2}$ ?

(spoiler)

Although $π$ looks like a variable, it’s a constant ( $\approx 3.14$ ). Therefore,

$\frac{d}{d x} (π^{2}) = 0$

Negative exponents

Use the power rule to differentiate $f (x) = x^{- 4}$ .

(spoiler)

$\frac{d}{d x} (x^{- 4})$

$= - 4 x^{- 4 - 1}$

$= - 4 x^{- 5}$

$= - \frac{4}{x ^{5}}$

AP tip:

Fractional exponents

Use the power rule to differentiate $f (x) = x^{1/2}$ .

(spoiler)

$\frac{d}{d x} (x^{1/2})$

$= \frac{1}{2} x^{(1/2) - 1}$

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

$= \frac{1}{2 x ^{1/2}}$

$= \frac{1}{2 x}$

AP tip:

Whenever you see a radical function, rewrite it using fractional exponents so you can apply the power rule. For example, $3 x = x^{1/3}$ and $4 x^{3} = x^{3/4}$ .

Try to simplify expressions when you can. For example, the expression

$\frac{x ^{3} + 3 x}{2 x}$

can be split into

$\frac{x ^{3}}{3 x} + \frac{3 x}{2 x}$

and simplified into

$\frac{1}{3} x^{2} + \frac{3}{2}$

Later on, you’ll learn how to differentiate quotients directly, but sometimes rewriting and simplifying lets you use the power rule instead.

::: sidenote When the power rule can’t be used Keep in mind the power rule only applies to functions where the exponent is a constant, not a variable. For example:

$\frac{d}{d x} (2^{x}) \neq = x \cdot 2^{x - 1}$

because $2^{x}$ is an exponential function, not a power function. :::

The correct derivative will be covered in section 3.5 (Special derivatives), where you’ll learn how to handle derivatives of exponentials and other functions.

AP tip:

Challenge problems

Find the derivative of $f (x) = \frac{1}{3 x ^{2}} + 2 x - 1$

Solution

(spoiler)

Answer: $f^{'} (x) = - \frac{2}{3 x ^{5/3}} + \frac{1}{x ^{1/2}}$

Rewrite the radical terms using fractional exponents.

$\frac{1}{3 x ^{2}} = \frac{1}{x ^{2/3}} = x^{- 2/3}$

and

$2 x = 2 x^{1/2}$

Now the function is

$f (x) = x^{- 2/3} + 2 x^{1/2} - 1$

Apply the power rule to each term.

Find the value(s) of $k$ such that $y = 3 x + k$ is a tangent line to $f (x) = x^{3}$ .

Hint: At the point of tangency, a function and its tangent line share:

The $y$ -value

The slope value

Solution

(spoiler)

Answer: $k = 2 or - 2$

It helps to sketch these two functions to see what the question is asking.

Using the power rule, the derivative is $f^{'} (x) = 3 x^{2}$ .

At a point of tangency $(a, f (a))$ , the function has the same slope as the tangent line.

The line $y = 3 x + k$ has slope $3$ . The function has slope $3$ when $f^{'} (a) = 3$ .

$3 a^{2} = 3$

$a^{2} = 1$

$a = \pm 1$

So the points of tangency are $(1, f (1)) = (1, 1)$ and $(- 1, f (- 1)) = (- 1, - 1)$ .

At each tangency point, the line and the function also share the same $y$ -value, so substitute $x = a$ into $y = 3 x + k$ and match it to $f (a)$ .

For $(1, 1)$ :

$3 (1) + k = (1)^{3}$

$k = - 2$

For $(- 1, - 1)$ :

$3 (- 1) + k = (- 1)^{3}$

$k = 2$

Find the derivative of $y = (x + 1) (x^{3} - 2)$

Solution

(spoiler)

Answer: $y^{'} = 4 x^{3} + 3 x^{2} - 2$

First, expand so the function is written as a sum of power terms.

$y = x (x^{3}) + x (- 2) + 1 (x^{3}) + 1 (- 2)$

$= x^{4} - 2 x + x^{3} - 2$

Next, apply the power rule to each term.

$y^{'} = 4 x^{3} - 2 + 3 x^{2} - 0$

$= 4 x^{3} + 3 x^{2} - 2$

In the next section, you’ll learn how to do this same problem using the product rule for the product of two functions.