Power rule | AP Calc AB Derivative basics

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, you can use shortcuts called 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 exponent.

Sidenote

A note on notation

The derivative of $y = f (x)$ can be written in several ways

$f^{'} (x)$

Lagrange’s notation, commonly used for derivatives of functions.

$\frac{d y}{d x}$

Leibniz’s notation, representing the rate of change of $y$ with respect to $x$ . It emphasizes that $y$ changes as $x$ changes. Later, variables other than $x$ and $y$ will be used.

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

The derivative operator. It acts on an expression like a verb, telling you to differentiate an expression.

For example, $\frac{d}{d x} [2 x]$ means “take the derivative of $2 x$ .”

The power rule

Use to differentiate algebraic functions of the form $x^{n}$ , where $n$ is any real number.

Power rule:

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

To differentiate $x^{n}$ :

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

For example,

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

Additional properties

Most expressions contain multiple terms or coefficients. These can be handled using specific differentiation properties.

Constant multiple rule:

Constant multiples can be “pulled out” of the derivative, evaluated, and multiplied back in at the end.

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

where $c$ is a real number.

Example 1.

$\frac{d}{d x} (- 5 x^{2}) = - 5 \cdot \frac{d}{d x} (x^{2}) = - 5 \cdot (2 x^{2 - 1}) = - 10 x$

Example 2.

$\frac{d}{d x} [(2 x)^{3}] = \frac{d}{d x} (8 x^{3}) = 8 \cdot \frac{d}{d x} (x^{3}) = 8 \cdot 3 x^{3 - 1} = 24 x^{2}$

Sum/difference rule:

Differentiate multi-term expressions by taking the derivative of each term individually.

$\frac{d}{d x} [f (x) \pm g (x)] = f^{'} (x) \pm g^{'} (x)$

Find the derivative of $f (x) = x^{4} - 3 x^{2}$ .

(spoiler)

Apply the power rule to each term separately:

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

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

1. Linear terms

Differentiate $f (x) = 4 x$ (note that $x$ can be considered $x^{1}$ ).

(spoiler)

$\frac{d}{d x} (4 x^{1}) = 4 \cdot 1 \cdot x^{1 - 1} = 4 x^{0} = 4$

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.

2. Constant terms

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)

$\frac{d}{d x} (2) = \frac{d}{d x} (2 x^{0}) = 2 \cdot 0 \cdot x^{0 - 1} = 0$

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

Find $g^{'} (x)$ for $g (x) = π^{2}$ .

(spoiler)

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

$\frac{d}{d x} (π^{2}) = 0$

3. Negative exponents

Use the power rule to differentiate $f (x) = - 2 x^{- 4}$ .

(spoiler)

The rule works the same:

$\frac{d}{d x} (- 2 x^{- 4}) = - 2 (- 4 x^{- 4 - 1}) = 8 x^{- 5} = \frac{8}{x ^{5}}$

AP tip:

Try to spot functions that are quotients with a number on top and an expression with a power on the bottom. These can be rewritten in the form $x^{n}$ (where $n$ is negative).

For example,

$\frac{2}{5 x ^{3}} = \frac{2}{5} x^{- 3}$

Once rewritten, the power rule can be applied directly.

4. Fractional exponents

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

(spoiler)

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

AP tip:

Radical functions can be rewritten using fractional exponents. For example,

$3 x = x^{1/3} 4 x^{3} = x^{3/4}$

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

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

can be split into

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

and simplified into

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

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 of exponential functions will be covered in section 3.5 (Special derivatives).

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, in $f (x) = x^{e}$ , the exponent $e$ is a constant. So the derivative is $f^{'} (x) = e \cdot x^{e - 1}$ .

Challenge problems

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

(spoiler)

First, rewrite the radical terms using fractional exponents. Rewritten,

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

Then, applying the power rule to each term,

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

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 have:

The same slope

The same $y$ -value

(spoiler)

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

Using the power rule, the derivative of $f$ is

$f^{'} (x) = 3 x^{2}$

At a point of tangency $(a, f (a))$ , the function and its tangent line have the same slope.

The tangent line $y = 3 x + k$ has a slope of $3$ , so $f^{'} (a) = 3$ as well. Then

$3 a^{2} = 3 a^{2} = 1 a = \pm 1$

At each point of tangency, the line and the curve also pass through the same point.

For $a = 1$ :

The point on the curve is $(1, f (1)) = (1, 1)$ . Since this point lies on the line $y = 3 x + k$ , we substitute $x = 1$ and $y = 1$ to find $k$ :

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

For $a = - 1$ :

The point on the curve is $(- 1, f (- 1)) = (- 1, - 1)$ . Substituting into $y = 3 x + k$ ,

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

Find the derivative of

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

(spoiler)

First, expand the polynomial into a sum of power terms.

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

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.

What you’ll learn

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

Sidenote

A note on notation

The derivative of $y = f (x)$ can be written in several ways

$f^{'} (x)$

Lagrange’s notation, commonly used for derivatives of functions.

$\frac{d y}{d x}$

Leibniz’s notation, representing the rate of change of $y$ with respect to $x$ . It emphasizes that $y$ changes as $x$ changes. Later, variables other than $x$ and $y$ will be used.

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

The derivative operator. It acts on an expression like a verb, telling you to differentiate an expression.

For example, $\frac{d}{d x} [2 x]$ means “take the derivative of $2 x$ .”

The power rule

Use to differentiate algebraic functions of the form $x^{n}$ , where $n$ is any real number.

Power rule:

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

To differentiate $x^{n}$ :

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

For example,

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

Additional properties

Most expressions contain multiple terms or coefficients. These can be handled using specific differentiation properties.

Constant multiple rule:

Constant multiples can be “pulled out” of the derivative, evaluated, and multiplied back in at the end.

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

where $c$ is a real number.

Example 1.

$\frac{d}{d x} (- 5 x^{2}) = - 5 \cdot \frac{d}{d x} (x^{2}) = - 5 \cdot (2 x^{2 - 1}) = - 10 x$

Example 2.

$\frac{d}{d x} [(2 x)^{3}] = \frac{d}{d x} (8 x^{3}) = 8 \cdot \frac{d}{d x} (x^{3}) = 8 \cdot 3 x^{3 - 1} = 24 x^{2}$

Sum/difference rule:

Differentiate multi-term expressions by taking the derivative of each term individually.

$\frac{d}{d x} [f (x) \pm g (x)] = f^{'} (x) \pm g^{'} (x)$

Find the derivative of $f (x) = x^{4} - 3 x^{2}$ .

(spoiler)

Apply the power rule to each term separately:

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

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

1. Linear terms

Differentiate $f (x) = 4 x$ (note that $x$ can be considered $x^{1}$ ).

(spoiler)

$\frac{d}{d x} (4 x^{1}) = 4 \cdot 1 \cdot x^{1 - 1} = 4 x^{0} = 4$

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.

2. Constant terms

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)

$\frac{d}{d x} (2) = \frac{d}{d x} (2 x^{0}) = 2 \cdot 0 \cdot x^{0 - 1} = 0$

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

Find $g^{'} (x)$ for $g (x) = π^{2}$ .

(spoiler)

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

$\frac{d}{d x} (π^{2}) = 0$

3. Negative exponents

Use the power rule to differentiate $f (x) = - 2 x^{- 4}$ .

(spoiler)

The rule works the same:

$\frac{d}{d x} (- 2 x^{- 4}) = - 2 (- 4 x^{- 4 - 1}) = 8 x^{- 5} = \frac{8}{x ^{5}}$

AP tip:

Try to spot functions that are quotients with a number on top and an expression with a power on the bottom. These can be rewritten in the form $x^{n}$ (where $n$ is negative).

For example,

$\frac{2}{5 x ^{3}} = \frac{2}{5} x^{- 3}$

Once rewritten, the power rule can be applied directly.

4. Fractional exponents

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

(spoiler)

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

AP tip:

Radical functions can be rewritten using fractional exponents. For example,

$3 x = x^{1/3} 4 x^{3} = x^{3/4}$

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

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

can be split into

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

and simplified into

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

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 of exponential functions will be covered in section 3.5 (Special derivatives).

AP tip:

Challenge problems

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

(spoiler)

First, rewrite the radical terms using fractional exponents. Rewritten,

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

Then, applying the power rule to each term,

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

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 have:

The same slope

The same $y$ -value

(spoiler)

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

Using the power rule, the derivative of $f$ is

$f^{'} (x) = 3 x^{2}$

At a point of tangency $(a, f (a))$ , the function and its tangent line have the same slope.

The tangent line $y = 3 x + k$ has a slope of $3$ , so $f^{'} (a) = 3$ as well. Then

$3 a^{2} = 3 a^{2} = 1 a = \pm 1$

At each point of tangency, the line and the curve also pass through the same point.

For $a = 1$ :

The point on the curve is $(1, f (1)) = (1, 1)$ . Since this point lies on the line $y = 3 x + k$ , we substitute $x = 1$ and $y = 1$ to find $k$ :

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

For $a = - 1$ :

The point on the curve is $(- 1, f (- 1)) = (- 1, - 1)$ . Substituting into $y = 3 x + k$ ,

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

Find the derivative of

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

(spoiler)

First, expand the polynomial into a sum of power terms.

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

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.