Derivation exercises that involve the variables or functions raised to a numerical exponent can be solved using the power rule formula. This formula allows us to derive variables such as but not limited to $latex x^n$, where $latex n$ is either a positive, negative or rational real number.

In special cases of functions such as polynomial and transcendental functions raised to a numerical exponent, the power rule is supported by another derivative rule. Here, we will look at the summary of the power rule. Additionally, we will explore several examples with answers to understand the application of the power rule formula.

## Summary of The Power Rule

The power rule is a very helpful tool to derive a variable or a function raised to a numerical exponent. It is a rule that 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 to one. This gives us the power rule formula in different cases such as:

### A) Single variable *x* raised to a numerical exponent *n*

The formula is:

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

You **can use** this form of the power rule formula to derive functions such as and limited to:

✔️ $latex y = x^n$

✔️ $latex y = \frac{1}{x^n} = x^{-n}$

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

where

$latex n = $ any real number

$latex \frac{n_1}{n_2} = n$, *for radical functions only*

and **not** in functions:

❌ $latex y = x^x$

❌ $latex y = x^{f(x)}$

### B) 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. Each algebraic term of the polynomial will use the basic power rule formula.

Then, sum/difference of derivatives will be applied to the whole polynomial function. To 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 highest degree of exponent 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) Transcendental Functions raised to a numerical exponent *n*

Transcendental functions are functions that cannot be expressed as a finite combination of the algebraic arithmetic operations of addition, subtraction, multiplication, or division. Some examples of these functions 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

$latex f'(u^n) = nu^{n-1} \cdot u’$ |

Where

- $latex u =$ any transcendental function
- $latex u’ =$ the derivative method of the transcendental function
- $latex n =$ the numerical exponent of the transcendental function

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## Power Rule – Examples with answers

Using the formula detailed above, we can derive various variables, polynomials, or transcendental functions raised to a numerical exponent. Each of the following examples has its respective detailed solution.

**EXAMPLE 1**

Find the derivative of $latex f(x) = x^{12}$.

##### Solution

The first thing we need to do is to identify the case and list the appropriate form of the power rule formula. Since this is a single variable raised to a numerical exponent, we can list down this form of power rule formula for our reference:

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

Then, let’s determine the exponent of our variable. In this case, our exponent is 12. Hence,

$latex n = 12$

We can now apply the power rule formula to derive the problem:

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

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

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

Simplifying algebraically, we have

$$\frac{d}{dx} (x^n) = 12 \cdot x^{11}$$

**And the final answer is:**

$latex f'(x) = 12x^{11}$

**EXAMPLE 2**

What is the derivative of $latex f(x) = x^{10}-5x^6+2x^5-3x^2+10$?

##### Solution

First and foremost, we need to identify the case and list the appropriate form of the power rule formula. Since this is a polynomial with different algebraic terms raised to different numerical exponents, we can list down this form of power rule formula for our reference:

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

We can now apply the power rule formula to derive the problem:

$$\frac{d}{dx} (x^{n_k} + … + x^{n_2} + x^{n_1} + c) =\frac{d}{dx} (x^{10}-5x^6+2x^5-3x^2+10)$$

$$= 10x^{10-1} – 5 \cdot (6x^{6-1}) + 2 \cdot (5x^{5-1})– 3 \cdot (2x^{2-1}) + 0$$

Simplifying algebraically, we have

$$\frac{d}{dx} (f(x)) = 10x^9 – 5(6x^5) + 2(5x^4) – 3(2x)$$

$$= 10x^9 – 30x^5 + 10x^4 – 6x$$

**And the final answer is:**

$$f'(x) = 10x^9 – 30x^5 + 10x^4 – 6x$$

**EXAMPLE 3**

Find the derivative of $latex f(x) = \frac{3}{x^{15}}$.

##### Solution

Let’s first identify the case and list the appropriate form of the power rule formula. Since this is a simple rational function, we can apply the laws of exponents to transform the rational form into its exponential form.

By doing this, we will have a single variable raised to a negative numerical exponent. Then, list down this form of power rule formula for our reference:

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

Let’s now convert the function from rational to exponential form by applying the laws of exponents:

$latex f(x) = \frac{3}{x^{15}}$

$latex f(x) = 3x^{-15}$

We can now apply the power rule formula to derive the problem:

$$\frac{d}{dx} (x^n) = \frac{d}{dx} (3x^{-15})$$

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

$$\frac{d}{dx} (x^n) = 3 \cdot (-15x^{(-15)-1})$$

Simplifying algebraically, we have

$$\frac{d}{dx} (x^n) = -45x^{-16}$$

Bringing the derived equation back into the rational form by applying the laws of exponents, we have

$$\frac{d}{dx} (x^n) = \frac{-45}{x^{16}}$$

**And the final answer is:**

$$f'(x) = -\frac{45}{x^{16}}$$

**EXAMPLE 4**

What is the derivative of $latex f(x) = 7 \sqrt[29]{x^{11}}$?

##### Solution

Since this is a radical function, we can apply the laws of exponents to transform the radical form into its exponential form. By doing this, we will have a single variable raised to a rational numerical exponent.

Then, list down this form of power rule formula for our reference:

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

Let us now convert the function from radical to exponential form:

$latex f(x) = 7 \sqrt[29]{x^{11}}$

$latex f(x) = 7 x^{\frac{11}{29}}$

Then, let’s determine the exponent of our variable. In this case, our exponent is $latex \frac{11}{29}$. Hence,

We can now apply the power rule formula to derive the problem:

$$\frac{d}{dx} (x^n) = \frac{d}{dx} (7x^{\frac{11}{29}})$$

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

$$ \frac{d}{dx} (x^n) = 7 \cdot \left[ \left(\frac{11}{29} \right) \cdot x^{\left(\frac{11}{29} \right)-1} \right]$$

Simplifying algebraically, we have

$$\frac{d}{dx} (x^n) = 7 \cdot \left(\frac{11}{29} x^{-\frac{18}{29}} \right)$$

$$\frac{d}{dx} (x^n) = \frac{77}{29} x^{-\frac{18}{29}} $$

Applying the laws of exponents, we have

$$\frac{d}{dx} (x^n) = \frac{77}{29x^{\frac{18}{29}}}$$

**And the final answer is:**

$$f'(x) = \frac{77}{29 \hspace{2.3 pt} \sqrt[29]{x^{18}}}$$*in radical form*

**EXAMPLE 5**

Find the derivative of $latex f'(x) = \frac{1}{\sqrt{x^5}}$

##### Solution

Since this is a hybrid of rational and radical functions, we can apply the laws of exponents to transform this function into its exponential form.

By doing this, we will have a single variable raised to a negative rational numerical exponent. Then, list down this form of power rule formula for our reference:

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

Let’s now convert the function from radical to exponential form:

$latex f(x) = \frac{1}{\sqrt{x^5}}$

$latex f(x) = x^{-\frac{5}{2}}$

Then, let’s determine the exponent of our variable. In this case, our exponent is $latex -\frac{5}{2}$. Hence,

We can now apply the power rule formula to derive the problem:

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

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

$$\frac{d}{dx} (x^n) = \left(-\frac{5}{2} \right) \cdot x^{\left(-\frac{5}{2} \right)-1}$$

Simplifying algebraically, we have

$$\frac{d}{dx} (x^n) = -\frac{5}{2} x^{-\frac{7}{2}}$$

Applying the laws of exponents, we have

$$\frac{d}{dx} (x^n) = -\frac{5}{2x^{\frac{7}{2}}}$$

**And the final answer is:**

$$f'(x) = -\frac{5}{2 \hspace{2.3 pt} \sqrt{x^7}}$$*in radical form*

**EXAMPLE 6**

What is the derivative of $latex f(x) = \sin^{2}{(x)}$?

##### Solution

Since this is a transcendental function, specifically a trigonometric function raised to a numerical exponent, we can list down this form of power rule formula for our reference:

$latex f'(u^n) = nu^{n-1} \cdot u’$

Let’s identify the transcendental function and the numerical exponent from the given problem:

Let

$latex u = \sin{x}$

$latex n = 2$

We can now apply the power rule formula to derive the problem:

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

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

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

Simplifying algebraically and applying applicable trigonometric identities, we have

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

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

**And the final answer is:**

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

**EXAMPLE 7**

Find the derivative of $latex f(x) = \frac{1}{e^{2x}}$.

##### Solution

Since this is a hybrid of rational and transcendental functions, we can apply the laws of exponents to transform this function into its exponential form.

By doing this, we will have a transcendental function raised to a negative numerical exponent. Then, we can list down this form of power rule formula for our reference:

$latex f'(u^n) = nu^{n-1} \cdot u’$

Let’s now convert the function from rational to exponential form:

$latex f(x) = \frac{1}{e^{2x}}$

$latex f(x) = e^{-2x}$

Since our transcendental function in this given problem is an exponential function, we can accept a variable exponent as part of the function’s characteristics. We can expand $latex e^{-2x}$ by applying the laws of exponents again:

$latex f(x) = e^{-2x}$

$latex f(x) = (e^x)^{-2}$

Then, let’s identify the transcendental function and the numerical exponent from the given problem:

Let

$latex u = e^x$

$latex n = -2$

We can now apply the power rule formula to derive the problem:

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

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

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

Simplifying algebraically and applying the laws of exponents, we have

$$\frac{d}{dx} (u^n) = 2 \cdot (e^x)^{-3} \cdot e^x$$

$$\frac{d}{dx} (u^n) = 2(e^x)^{-3+1}$$

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

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

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

**And the final answer is:**

$$f'(x) = \frac{2}{e^{2x}}$$

## Power Rule – Practice problems

Solve the following derivation 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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