Power rule | Derivative basics | Achievable AP Calculus AB

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 having to use the limit definition every time to find a derivative (which can get tricky with more complicated functions), there are shortcuts, or derivative rules, to do so quickly. One of the most useful is the power rule, for functions with a variable raised to a 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 $n$ and 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 encounter won’t be as simple as a single $x^{n}$ term. Fortunately, there are additional rules that make it easy 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.

Because multiplication is commutative, multiplying the constant by the derivative doesn’t change the differentiation process.

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 several added or subtracted terms, just differentiate each term individually.

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 applies to functions with terms that are linear, constants, or have negative or fractional exponents.

1. Linear terms

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

(spoiler)

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

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

$= 4 x^{0}$

$= 4$

In other words, the derivative of a linear term is just the coefficient that is in front of it, which should make sense - the tangent line to any point on a straight line like $y = 4 x$ will match the line itself, and have the same slope everywhere.

2. When f(x) = a constant

The derivative of a constant is always $0$ . The power rule is applied by noting that although $x$ is out of sight, $x^{0} = 1$ :

Differentiate $f (x) = 2$ .

(spoiler)

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

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

$= 0 x^{0 - 1}$

$= 0$

Hopefully this also makes sense intuitively, since the original function is a horizontal line with slope $0$ .

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

(spoiler)

Although $π$ looks like a symbol or a variable, don’t forget that it’s just a constant $\approx 3.14$ . Therefore

$\frac{d}{d x} (π^{2}) = 0$

3. 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 also be turned into the form $x^{n}$ ( $n$ will be a negative exponent). For example, $\frac{2}{x ^{3}}$ must become $2 x^{- 3}$ before the power rule can be applied.

4. Fractional exponents

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

(spoiler)

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

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

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

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

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

AP tip:

Whenever you see a radical function, turn it into an expression with fractional exponents to apply the power rule. E.g. $3 x = x^{1/3}, 4 x^{3} = x^{3/4}$ , etc.

Try to simplify expressions where possible. An expression such as

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

can be split up 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 calculate derivatives of functions expressed as a quotient, but sometimes recognizing that the simpler power rule can be used instead will make calculations easier.

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 and 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 $e \approx 2.72$ ) and not variables, so the power rule CAN be applied if those are in the exponent. For example, in $f (x) = x^{e}$ , $e$ is a constant in the exponent, and the derivative is $f^{'} (x) = e \cdot x^{e - 1}$ .

Challenge problems

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

Solution

(spoiler)

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

1. The term with the radical $\frac{1}{3 x ^{2}}$ can be turned into

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

as can the term

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

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

3. Next, the power rule can be applied to each term.

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

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’s helpful to draw out a general picture of these two functions for a rough idea of what the question is asking for.

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

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

$y = 3 x + k$ has a slope of $3$ . The derivative $3 x^{2}$ also has slope $3$ at some point $x = a$ . Solving,

$3 a^{2} = 3$

$a^{2} = 1$

$a = \pm 1$

Then the points of tangency $(a, f (a))$ are $(1, 1)$ or $(- 1, - 1)$ .

Additionally, the function and the tangent line share the same $y$ -value at each point of tangency, or $3 x + k = x^{3}$ at the specific points $x = a$ .

For $(1, 1)$ :

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

For $(- 1, - 1) :$

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

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

Solution

(spoiler)

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

First, expand so that the function has several terms containing powers.

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

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

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.

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 $n$ and subtract $1$ from the exponent.

For example,

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

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

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.

Because multiplication is commutative, multiplying the constant by the derivative doesn’t change the differentiation process.

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 several added or subtracted terms, just differentiate each term individually.

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 applies to functions with terms that are linear, constants, or have negative or fractional exponents.

1. Linear terms

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

(spoiler)

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

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

$= 4 x^{0}$

$= 4$

2. When f(x) = a constant

The derivative of a constant is always $0$ . The power rule is applied by noting that although $x$ is out of sight, $x^{0} = 1$ :

Differentiate $f (x) = 2$ .

(spoiler)

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

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

$= 0 x^{0 - 1}$

$= 0$

Hopefully this also makes sense intuitively, since the original function is a horizontal line with slope $0$ .

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

(spoiler)

Although $π$ looks like a symbol or a variable, don’t forget that it’s just a constant $\approx 3.14$ . Therefore

$\frac{d}{d x} (π^{2}) = 0$

3. 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:

4. Fractional exponents

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

(spoiler)

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

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

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

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

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

AP tip:

Whenever you see a radical function, turn it into an expression with fractional exponents to apply the power rule. E.g. $3 x = x^{1/3}, 4 x^{3} = x^{3/4}$ , etc.

Try to simplify expressions where possible. An expression such as

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

can be split up 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 calculate derivatives of functions expressed as a quotient, but sometimes recognizing that the simpler power rule can be used instead will make calculations easier.

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 and 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

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

Solution

(spoiler)

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

1. The term with the radical $\frac{1}{3 x ^{2}}$ can be turned into

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

as can the term

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

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

3. Next, the power rule can be applied to each term.

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’s helpful to draw out a general picture of these two functions for a rough idea of what the question is asking for.

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

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

$y = 3 x + k$ has a slope of $3$ . The derivative $3 x^{2}$ also has slope $3$ at some point $x = a$ . Solving,

$3 a^{2} = 3$

$a^{2} = 1$

$a = \pm 1$

Then the points of tangency $(a, f (a))$ are $(1, 1)$ or $(- 1, - 1)$ .

Additionally, the function and the tangent line share the same $y$ -value at each point of tangency, or $3 x + k = x^{3}$ at the specific points $x = a$ .

For $(1, 1)$ :

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

For $(- 1, - 1) :$

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

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

Solution

(spoiler)

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

First, expand so that the function has several terms containing powers.

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

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

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.