The **Power Rule** is one of the major and most commonly used formulas in Differential Calculus (*or Calculus I*). It is commonly applied in deriving a single variable, a set of polynomials, or a function with a numerical exponent. The power rule can be proven and developed by applying the binomial theorem through limits. Logarithmic differentiation can also be used to prove the power rule.

In this article, we will discuss everything about the power rule. We will cover its definition, formula, proofs, and application usage. We will also look at some examples and practice problems to apply the principles of the power rule.

## The Power Rule and its formula

### What is the Power Rule?

The power rule states that the derivative of a variable raised to a numerical exponent is equal to the value of the numerical exponent multiplied by the variable raised to the quantity of the numerical exponent subtracted by one.

Other forms and cases of the power rule, such as the case for polynomials and functions raised to a numerical exponent or the general power rule formula are also used for efficiency and convenience.

These special cases of the power rule are dependent on the principles of other derivative rules such as the sum/difference of derivatives for polynomials or the chain rule for functions raised to a numerical exponent.

Let’s begin with the basic form of the power rule.

### A) The basic Power Rule formula for a single variable raised to a numerical exponent *n*

The Power Rule Formula is:

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

where

- $latex n =$ the numerical value of the exponent limited to real numbers only
- $latex x =$ the variable that is being raised to a numerical exponent $latex n$

and $latex \frac{d}{dx}(x^n)$ can also be denoted by $latex y’$, $latex F'(x)$, $latex f'(x)$, or other letters used to denote functions with the apostrophe symbol.

## Special cases and forms of the Power Rule formula

### B) The Power Rule in polynomials

Polynomial functions are the sum/difference of algebraic terms with different exponents. Thus, the power rule formula to be used in polynomial functions will be supported by the sum/difference of derivatives.

In this special case, the power rule is stated as the derivative of a polynomial with algebraic terms of different exponents, which is equal to the summation of the derivatives of each term.

Each term will be derived either by using the basic power rule formula for all terms with variables or the derivative of constants in case the last term is a constant. To better illustrate, the formula is:

$latex f'(x^{n_k} + … + x^{n_2} + x^{n_1} + c) = {n_k} x^{{n_k}-1} + … + {n_2} x^{{n_2}-1} + {n_1} x^{{n_1}-1} + 0$ |

Where

- $latex n_k$ = the exponent of the algebraic term with the exponent of the highest degree in the polynomial
- $latex n_{\texttt{\#}}$ = the exponents of the other algebraic terms in the polynomial
- $latex c$ = constant, which if derived, is equal to zero

### C) The Power Rule in quantities of functions and transcendental functions raised to a numerical exponent *n*

This special case of the Power Rule, commonly called “The General Power Rule Formula”, is often confused to be similar to The Chain Rule Formula. However, this can also be considered as a special case of The Chain Rule Formula as both of these two rules depend on each other’s principles when deriving a transcendental function that is raised to a numerical exponent $latex n$.

For a brief recap, transcendental functions are functions that cannot be expressed as a finite combination of the algebraic arithmetic operations of addition, subtraction, multiplication, or division. Examples of these are trigonometric functions, logarithmic functions, their inverse functions if they exist, etc.

When raised to a numerical exponent $latex n$, the power rule is applied with the chain rule formula. The power rule is used as the derivative of the outside function *f* of the composite function $latex f(g(x))$. To illustrate, the formula is

$latex f'(u^n) = n(u)^{n-1} \cdot u’$ |

Where

- $latex u$ can either be a quantity of functions or a transcendental function
- $latex u’ =$ the derivative of the quantity of functions or the derivative method of the transcendental function
- $latex n =$ the numerical exponent of the quantity of functions or the transcendental function

You can visit our article Chain Rule Formula, Proof, and Examples, to know more about the differences between power and chain rules.

Start now: Explore our additional Mathematics resources

## Proofs of The Power Rule

### Proof of The Power Rule Using The Binomial Theorem

The power rule can be proved and derived using the binomial theorem. We can then obtain an expression using the binomial theorem and substitute it into the limit definition of a derivative.

By manipulating and simplifying the expression, we can prove the power rule. You can look at a detailed proof using the binomial theorem in this article.

### Proof of The Power Rule Using Logarithmic Differentiation

Unlike the proof of the power rule using the binomial theorem, proving this rule by logarithmic differentiation is actually more straightforward and the shortest method of proving the power rule formula considering you have foundations and learnings logarithmic differentiation.

You can look at our article about the Proofs of The Power Rule to learn how we can prove the power rule using logarithmic differentiation.

## When to use the Power Rule formula

### A) To derive a single variable raised to a numerical exponent *n*

This form of Power Rule Formula

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

**can be used** for the following conditions:

✔️ Raised to a positive numerical exponent:

$latex y = x^n$

where $latex x$ is a variable and $latex n$

is the positive numerical exponent

✔️ Raised to a negative exponent (rational function in exponential form):

$latex y = \frac{1}{x^n}$

$latex y = x^{-n}$

where $latex x$ is a variable and $latex n$

is the negative numerical exponent

✔️ Raised to a rational exponent (radical function in exponential form):

$latex y = \sqrt[n_2]{x^{n_1}}$

$latex y = x^{\frac{n_1}{n_2}}$

where $latex x$ is a variable and $latex \frac{n_1}{n_2} = n$

or the rational numerical exponent $latex n$

and **cannot be used** to derive:

❌ Raised to a variable exponent:

$latex y = x^x$

❌ Raised to any type of function:

$latex y = x^{f(x)}$

### B) Polynomials

This form of Power Rule Formula

$$f'(x^{n_k} + … + x^{n_2} + x^{n_1} + c)= {n_k} x^{{n_k}-1} + … + {n_2} x^{{n_2}-1} + {n_1} x^{{n_1}-1} + 0$$

**can be used** for the following conditions such as but not limited to:

✔️ $latex y = x^{n_k} + … + x^{n_2} + x^{n_1} + c$

or for instance,

✔️ $latex y = x^{n_k} + … + x^3 + x^2 + x + c$

* as long as none* of the algebraic terms are raised to a variable or a function.

### C) Transcendental functions raised to a numerical exponent *n*

This form of the Power Rule formula

$latex f'(u^n) = n(u)^{n-1} \cdot u’$

**can be used** for the following conditions:

✔️ $latex y = u^n$

where $latex u$ can be a quantity of functions or a transcendental function and $latex n$ is its numerical exponent. Such examples are but not limited to:

✔️ $latex y = (x^{n_k} + … + x^{n_2} + x^{n_1} + c)^n$

✔️ $latex y = \sin^{n}{x}$

*✔️ $latex y = (\ln{x})^n$*

**All exponents** of any quantity of functions or transcendental function **must not** be a variable or any other type of function.

❌ $latex y = u^x$

❌ $latex y = u^{f(x)}$

In the case of * exponential functions*, which is also a transcendental function, an exponential function naturally carries a variable exponent as part of its characteristics. When a quantity of exponential function $latex a^x$ is raised to a numerical exponent $latex n$, the same power rule formula

**can be used**. Such forms are:

✔️ $latex y = (a^x)^n$

✔️ $latex y = a^{nx}$

where $latex a$ is a real number

However, a quantity of exponential function $latex a^x$ raised to another variable $latex x$ or any other type of function $latex f(x)$, **shall not** use the power rule formula. Such forms are:

❌ $latex y = (a^x)^x$

❌ $latex y = a^{x^2}$

where $latex a$ is a real number

## How to use the Power Rule to find derivatives, a step by step tutorial

We are asked to derive

$latex f(x) = x^2$

As you can observe, this given function is a variable raised to a power of 2. But in order to derive this problem, we can use the power rule as shown by the following steps:

**Step 1:** It is always recommended to first identify the case in order to determine the best appropriate form of the power rule formula we can use. In this case, it is a single variable raised to a numerical exponent.** Step 2:** List down the form of power rule formula appropriate for this case:

$latex f'(x^n) = nx^{n-1}$

** Step 3:** If the original form of the given problem is rational or radical, apply the laws of exponents and convert them into exponential form. In this case, we can skip this step since our given is already in exponential form.

** Step 4:** Determine the exponent of the variable. In this case, our exponent is 2. Hence,

$latex n = 2$

** Step 5:** Apply the power rule formula to derive the problem:

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

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

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

** Step 6:** Simplify algebraically and apply the necessary trigonometric identities as well as other functional rules to the derived equation whenever applicable.

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

$$\frac{d}{dx} (x^n) = 2x^{1}$$

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

** Step 7:** If you think the derived equation cannot be simplified any longer, declare it as your final derivative answer:

$latex f'(x) = 2x$

For formality purposes, it is recommended that you use either $latex f'(x), y’,$ or $latex \frac{d}{dx}(f(x))$ as your derivative symbol on the left hand side of the derived final answer instead of $latex (x^n)’$ or $latex \frac{d}{dx}(x^n)$.

## Power Rule – Examples with answers

Each of the following examples has its respective detailed solution. It is recommended for you to try to solve the sample problems yourself before looking at the solution so that you can practice and fully master this topic.

**EXAMPLE 1**

Derive: $latex f(x) = -12x^{-13}$.

##### Solution

** Step 1: **Let us first identify the case in order to determine the best appropriate form of the power rule formula we can use. In this case, it is a single variable raised to a numerical exponent.

**List down the form of power rule formula appropriate for this case:**

**Step 2:**$latex f'(x^n) = nx^{n-1}$

** Step 3:** Since the given problem is already in exponential form, we do not need to apply the law of exponents before deriving the problem.

** Step 4:** Determine the exponent of the variable. In this case, our exponent is -13. Hence,

$latex n = -13$

** Step 5:** Apply the power rule formula to derive the problem:

$$ \frac{d}{dx} (x^n) = \frac{d}{dx} (-12x^{-13})$$

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

$$\frac{d}{dx} (x^n) = -12 \cdot (-13 \cdot x^{-13-1})$$

** Step 6:** Simplify algebraically:

$$\frac{d}{dx} (x^n) = -12 \cdot (-13 \cdot x^{-13-1})$$

$$\frac{d}{dx} (x^n) = -12 \cdot (-13x^{-14})$$

$$\frac{d}{dx} (x^n) = 156x^{-14}$$

Since the exponent is negative, we can apply the laws of exponents to further simplify it into rational form, although this optional:

$$\frac{d}{dx} (x^n) = \frac{156}{x^{14}}$$

** Step 7:** If you think the derived equation cannot be simplified any longer, declare it as your final derivative answer:

$$f'(x) = \frac{156}{x^{14}}$$

**EXAMPLE 2**

Find the derivative of $latex f(x) = 6x^3-21x^2-100x+625$.

##### Solution

** Step 1: **In this case, we have a polynomial.

**List down the form of power rule formula appropriate for this case:**

**Step 2:**$$f'(x^{n_k} + … + x^{n_2} + x^{n_1} + c)$$

$$= {n_k} x^{{n_k}-1} + … + {n_2} x^{{n_2}-1} + {n_1} x^{{n_1}-1} + 0$$

** Step 3:** Since all terms in the given problem are already in exponential form, we do not need to apply the law of exponents before deriving the problem.

** Step 4:** Determine the exponent for each term of the polynomial. Let,

$$f(x) = 6x^{n_3} – 21x^{n_2} – 100x^{n_1} + 625$$

and if

$$f(x) = 6x^3-21x^2-100x+625$$

Then we have

$latex n_1 = 1$

$latex n_2 = 2$

$latex n_3 = 3$

** Step 5:** Apply the power rule formula to derive the problem:

$$\frac{d}{dx} (6x^3-21x^2-100x+625) = 6 \cdot {n_3} x^{{n_3}-1}– 21 \cdot {n_2} x^{{n_2}-1}- 100 \cdot {n_1} x^{{n_1}-1} + 0$$

$$= 6 \cdot 3x^{3-1} – 21 \cdot 2x^{2-1} – 100 \cdot 1x^{1-1} + 0$$

** Step 6:** Simplify algebraically:

$$6 \cdot 3x^{3-1} – 21 \cdot 2x^{2-1} – 100 \cdot 1x^{1-1} + 0=18x^2 – 42x – 100$$

** Step 7:** If you think the derived equation cannot be simplified any longer, declare it as your final derivative answer:

$latex f'(x) = 18x^2 – 42x – 100$

**EXAMPLE 3**

Derive $$f(x) = x^{12} – 3 \sqrt{x} + \frac{5}{x^8} – \frac{9}{\sqrt[3]{x^2}}$$

##### Solution

** Step 1: **In this case, we have a polynomial.

**List down the form of power rule formula appropriate for this case:**

**Step 2:**$$f'(x^{n_k} + … + x^{n_2} + x^{n_1} + c)$$

$$= {n_k} x^{{n_k}-1} + … + {n_2} x^{{n_2}-1} + {n_1} x^{{n_1}-1} + 0$$

** Step 3:** Some of the terms are not in exponential form, therefore, we need to apply the laws of exponents for the terms that are either in radical, rational, or both:

$$f(x) = x^{12} – 3 \sqrt{x} + \frac{5}{x^8} – \frac{9}{\sqrt[3]{x^2}}$$

$$ f(x) = x^{12} – 3x^{\frac{1}{2}} + 5x^{-8} – 9x^{-\frac{2}{3}}$$

** Step 4:** Determine the exponent for each term of the polynomial. Let,

$$f(x) = x^{n_4} – 3x^{n_3} + 5x^{n_2} – 9x^{n_1}$$

and if

$$ f(x) = x^{12} – 3x^{\frac{1}{2}} + 5x^{-8} – 9x^{-\frac{2}{3}}$$

Then we have

$latex n_1 = -\frac{2}{3}$

$latex n_2 = -8$

$latex n_3 = \frac{1}{2}$

$latex n_4 = 12$

** Step 5:** Apply the power rule formula to derive the problem:

$$x^{12} – 3x^{\frac{1}{2}} + 5x^{-8} – 9x^{-\frac{2}{3}} = {n_4}x^{n_4 – 1} – 3 \cdot ({n_3}x^{n_3 – 1}) + 5 \cdot ({n_2}x^{n_2 – 1}) – 9 \cdot ({n_1}x^{n_1 – 1})$$

$$= 12x^{12-1} – 3 \cdot \left(\frac{1}{2} x^{\frac{1}{2} – 1} \right) + 5 \cdot (-8x^{-8 – 1}) – 9 \cdot \left(-\frac{2}{3} x^{-\frac{2}{3} – 1} \right)$$

** Step 6:** Simplify algebraically:

$$12x^{12-1} – 3 \cdot \left(\frac{1}{2} x^{\frac{1}{2} – 1} \right) + 5 \cdot (-8x^{-8 – 1}) – 9 \cdot \left(-\frac{2}{3} x^{-\frac{2}{3} – 1} \right) = 12x^{11} – \frac{3}{2} x^{-\frac{1}{2}} – 40x^{-9} + 6x^{-\frac{5}{3}}$$

Since some of the terms have negative or rational exponents, we can apply the laws of exponent to further simplify:

$$12x^{11} – \frac{3}{2} x^{-\frac{1}{2}} – 40x^{-9} + 6x^{-\frac{5}{3}} = 12x^{11} – \frac{3}{2 \sqrt{x}} – \frac{40}{x^9} + \frac{6}{\sqrt[3]{x^5}}$$

** Step 7:** If you think the derived equation cannot be simplified any longer, declare it as your final derivative answer:

$$f'(x) = 12x^{11} – \frac{3}{2 \sqrt{x}} – \frac{40}{x^9} + \frac{6}{\sqrt[3]{x^5}}$$

**EXAMPLE 4**

What is the derivative of $latex f(x) = e^{12x}$?

##### Solution

** Step 1: **In this case, we have a transcendental function, specifically, an exponential function.

** Step 2:** List down the form of power rule formula appropriate for this case:

$latex f'(u^n) = nu^{n-1} \cdot u’$

** Step 3:** Although, the exponential function is already in exponential form, in the special case of this function, it is important to apply the laws of exponent to extract the exponent of the function:

$latex f(x) = e^{12x}$

$latex f(x) = (e^x)^{12}$

** Step 4:** Determine the exponent for the exponential function. In this case, our exponent is 12. Hence,

$latex n = 12$

** Step 5:** Apply the power rule formula to derive the problem:

$$\frac{d}{dx} (u^n) = \frac{d}{dx} ((e^x)^{12})$$

$$\frac{d}{dx} (u^n) = nu^{n-1} \cdot u’$$

$$\frac{d}{dx} (u^n) = 12 \cdot (e^x)^{12-1} \cdot e^x$$

** Step 6:** Simplify algebraically:

$$\frac{d}{dx} (u^n) = 12 \cdot (e^x)^{12-1} \cdot e^x$$

$$\frac{d}{dx} (u^n) = 12(e^x)^{12-1+1}$$

$$\frac{d}{dx} (u^n) = 12(e^x)^{12}$$

Applying the laws of exponents to further simplify, we have

$$\frac{d}{dx} (u^n) = 12e^{12x}$$

** Step 7:** If you think the derived equation cannot be simplified any longer, declare it as your final derivative answer:

$latex f'(x) = 12e^{12x}$

**EXAMPLE 5**

Derive: $latexf(x) = 2 \sinh^{2}{(x)}$.

##### Solution

** Step 1: **In this case, we have a transcendental function, specifically, a hyperbolic function.

** Step 2:** List down the form of the power rule formula appropriate for this case:

$latex f'(u^n) = nu^{n-1} \cdot u’$

** Step 3:** Although, the given function is already in exponential form, we can still rewrite it into quantity form if that would lessen the confusion.

$latex f(x) = 2 \sinh^{2}{(x)}$

$latex f(x) = 2(\sinh{(x)})^2$

** Step 4:** Determine the exponent for the hyperbolic function. In this case, our exponent is 2. Hence,

$latex n = 2$

** Step 5:** Apply the power rule formula to derive the problem:

$$\frac{d}{dx} (u^n) = \frac{d}{dx} (2(\sinh{(x)})^2)$$

$$ \frac{d}{dx} (u^n) = nu^{n-1} \cdot u’$$

$$\frac{d}{dx} (u^n) = 2 \cdot (2 \cdot (\sinh{(x)})^{2-1} \cdot \cosh{(x)})$$

** Step 6:** Simplify algebraically and apply applicable hyperbolic identities:

$$\frac{d}{dx} (u^n) = 2 \cdot (2 \cdot (\sinh{(x)})^{2-1} \cdot \cosh{(x)})$$

$$\frac{d}{dx} (u^n) = 2 \cdot (2 \sinh{(x)} \cosh{(x)})$$

$$\frac{d}{dx} (u^n) = 2 \sinh{(2x)}$$

** Step 7:** If you think the derived equation cannot be simplified any longer, declare it as your final derivative answer:

$latex f'(x) = 2 \sinh{(2x)}$

## Power Rule – Practice problems

Solve the following differentiation problems and test your knowledge on this topic. Use the power rule formula detailed above to solve the exercises. If you have problems with these exercises, you can study the examples solved above.

## See also

Interested in learning more about the power rule? Take a look at these pages:

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