banner

Power Rule – Derivation, Explanation, and Example

The power rule is one of the first many derivative rules you’ll learn in your differential calculus classes. Taking the derivative of expressions raised to a certain power can be tedious if we use the definition of derivative to differentiate it. Still, thanks to the power rule, this won’t be a problem for us anymore.

The power rule helps us find the derivative of functions and expressions raised to a power.

We’ll explore how this particular derivative rule was derived and understand why we need to establish a power rule for expressions with a higher power. Of course, we’ll try out different examples to make sure that we know this rule by heart at the end of this article.

What is the power rule?                

The power rule helps us determine the derivative of $f(x) = x^n$ by using the exponent as the new coefficient and decreasing the degree by $\boldsymbol{1}$. Before we dive into the process of differentiating $f(x) = x^n$, let’s recall the definition of derivative and see how we’ll derive the formula using what we know so far.

Deriving the power rule formula

We know that the derivative can be attained in your previous lessons by limiting the function’s difference quotient as $h$ approaches $0$.

\begin{aligned} \dfrac{dy}{dx} &= \lim_{h\rightarrow 0} \dfrac{f(x + h) – f(x)}{h}\end{aligned}

The motivation behind why the power rule was established is that $f(x + h)$ will be difficult to expand when working with higher powers.

\begin{aligned}(x+h)^1 &= x + h\\(x + h)^2 &= x^2 + 2xh + h^2\\(x +h)^3 &= x^3 + 3x^2h +3xh^2 + h^3\\.\\.\\.\end{aligned}

Yes, we can use Pascal’s triangle to expand binomial expressions faster. However, it will still take time, especially if we want to evaluate a long list of expressions with a power of $n$. This is when the power rule comes to the rescue!

Let’s stick with the fact that $(x + h)^n = x^n + n x^{n -1}h + a_2x^{n – 2}h^2 + …+ a_{n-1}xh^{n -1} + h^n$. Using the definition of derivatives, we have the following expression for $f’(x)$, where $f(x) = x^n$.

\begin{aligned}f(x)&=x^n\\f'(x) &= \lim_{h\rightarrow 0} \dfrac{f(x + h)- f(x)}{h}\\&= \lim_{h\rightarrow 0} \dfrac{(x^n + n x^{n -1}h + a_2x^{n -2}h^2 + …+ a_{n-1}xh^{n -1} + h^n) – x^n}{h}\\&= \lim_{h\rightarrow 0} \dfrac{h(n x^{n -1} + a_2x^{n -2}h + …+ a_{n-1}xh^{n -2} + h^{n-1})}{h}\\&= \lim_{h\rightarrow 0} n x^{n -1} + a_2x^{n -2}h + …+ a_{n-1}xh^{n -2} + h^{n-1}\\&= nx^{n-1} + 0 + …+ 0+0\\&= nx^{n -1} \end{aligned}

From this, we can see that $f’(x)  =nx^{n -1}$ and this is the power rule. This means that the power rule tells us that given $\boldsymbol{f(x) = x^n}$, the derivative of $\boldsymbol{f(x)}$ is simply equal to $\boldsymbol{nx^{n -1}}$.

How to use the power rule?                

Let’s repeat the power rule. If $f(x) = x^n$, then $f’(x) = x^{n -1}$. Now, here are some steps to guide you on how you can apply this amazing rule to differentiate any powered expressions:

  • Given $f(x) = x^n$, take note of $n$ and use this as the derivative’s new coefficient.
  • The new exponent of $f(x)$’s derivative is simply one degree lower than the previous exponent.

As an example, we can try evaluating the derivative of $f(x) = x^4$. We can use $4$ as the derivative’s coefficient then take the exponent down by $1$ for the derivative’s new degree.

\begin{aligned}f(x) &= x^4\\ f’(x) &= 4(x)^{4 -1}\\&= 4x^3 \end{aligned}

This shows that we can differentiate $f(x) = x^4$ in a few seconds through the power rule. In fact, we have $f’(x) = 4x^3$. As an added exercise for you to do on your own: try to derive this using the definition of derivatives. Once done, you’ll thank the power rule for saving you so much time!

Example 1

Evaluate the following expressions:

a. $ \dfrac{d}{dx} \left(x^{\displaystyle{6}} \right )$
b. $ \dfrac{d}{dx} \left(x^{\displaystyle{12}} \right )$
c. . $ \dfrac{d}{dx} \left(x^{\displaystyle{-5}} \right )$

Solution

When working with functions of the form $f(x) = x^n$, we always go back to the fact that $f’(x) = nx^{n -1}$.

Working on the first item, $ \dfrac{d}{dx} \left(x^{\displaystyle{6}} \right )$, we can use $6$ as the derivative’s coefficient and decrease the exponent by $1$ for the derivative of $x^{\displaystyle{6}}$.

\begin{aligned}\dfrac{d}{dx} \left(x^{\displaystyle{6}} \right ) &= 6 x^{\displaystyle{6 -1 }} \\&= 6 x^{\displaystyle{5}}\end{aligned}

We’ll apply a similar process for the two remaining items as shown below.

\begin{aligned}\dfrac{d}{dx} \left(x^{\displaystyle{12}} \right ) &= 12 x^{\displaystyle{12 -1 }} \\&= 12 x^{\displaystyle{11}}\end{aligned}

\begin{aligned}\dfrac{d}{dx} \left(x^{\displaystyle{-5}} \right ) &= -5 x^{\displaystyle{-5 -1 }} \\&= -5x^{\displaystyle{-6}}\\&= – \dfrac{5}{x^{\displaystyle{6}}}\phantom{xxx} ,\color{green}a^{\displaystyle{-m}} = \dfrac{1}{a^{\displaystyle{m}}}\end{aligned}

Example 2

Evaluate the following expressions:

a. $ \dfrac{d}{dx} \left(12 x^{\displaystyle{3}} \right )$
b. $ \dfrac{d}{dx} \left(\dfrac{1}{2} x^{\displaystyle{20}} \right )$
c. . $ \dfrac{d}{dx} \left(6 x^{\displaystyle{-1}} \right )$

Solution

We will continue to use the power rule, $f’(x) = nx^{n -1}$, for these three items. But do you notice something different about these expressions from the first? We now have coefficients before each expression.

When this happens, we can apply constant multiple rule, $\dfrac{d}{dx} cf(x) = c \cdot \dfrac{d}{dx} f(x)$, or $\dfrac{d}{dx} c f(x) = c f’(x)$. This means that we can simply multiply the coefficients to the resulting derivative.

Let’s work on the first item, $ \dfrac{d}{dx} \left(12 x^{\displaystyle{3}} \right )$, we have the solution shown below:

\begin{aligned}\dfrac{d}{dx} \left(12 x^{\displaystyle{3}} \right ) &= 12 \dfrac{d}{dx} \left( x^{\displaystyle{3}} \right ), \phantom{xxx} \color{green}\text{Constant Multiple Rule}\\&= 12 \left( 3x^{\displaystyle{3 – 1}} \right ), \phantom{xxx} \color{green}\text{Power Rule}\\&= 12 \left( 3x^{\displaystyle{2}} \right )\\&= 36x^{\displaystyle{2}}  \end{aligned}

We can use a similar process to find the derivative of the two remaining items as shown below.

\begin{aligned}\dfrac{d}{dx} \left(\dfrac{1}{2}x^{\displaystyle{20}} \right ) &= \dfrac{1}{2} \dfrac{d}{dx} \left( x^{\displaystyle{20}} \right ), \phantom{x} \color{green}\text{Constant Multiple Rule}\\&= \dfrac{1}{2} \left( 20x^{\displaystyle{20 – 1}} \right ), \phantom{x} \color{green}\text{Power Rule}\\&= \dfrac{1}{2} \left( 20x^{\displaystyle{19}} \right )\\&= 10x^{\displaystyle{19}}  \end{aligned}

\begin{aligned}\dfrac{d}{dx} \left(6x^{\displaystyle{-1}} \right ) &= 6 \dfrac{d}{dx} \left( x^{\displaystyle{-1}} \right ), \phantom{x} \color{green}\text{Constant Multiple Rule}\\&= 6 \left( -1x^{\displaystyle{-1 – 1}} \right ), \phantom{x} \color{green}\text{Power Rule}\\&= 6 \left( -x^{\displaystyle{-2}} \right )\\&= -6x^{\displaystyle{-2}}\\&= -\dfrac{6}{x^{\displaystyle{2}}}  \end{aligned}

Example 3

Evaluate the following expressions:

a. $ \dfrac{d}{dx} \left(\pi x^{\displaystyle{\pi}} \right )$
b. $ \dfrac{d}{dx} \sqrt[3]{x}$

Solution

The first item may appear tricky because of the appearance of $\pi$, but just keep in mind that it’s still a coefficient that we can work with like any other coefficients. Hence, we have the following:

 \begin{aligned}\dfrac{d}{dx} \left(\pi x^{\displaystyle{\pi}} \right ) &= \pi \dfrac{d}{dx} \left( x^{\displaystyle{3}} \right ), \phantom{xxx} \color{green}\text{Constant Multiple Rule}\\&= \pi \left( \pi x^{\displaystyle{\pi – 1}} \right ), \phantom{xxx} \color{green}\text{Power Rule}\\&= \pi \left( \pi x^{\displaystyle{\pi – 1}} \right )\\&= \pi^2 x^{\displaystyle{\pi – 1}}  \end{aligned}

At first glance, $ \dfrac{d}{dx} \sqrt[3]{x}$, does not seem to be an expression that will benefit from the power rule. If you express $\sqrt[3]{x}$ as a power of $x$, we have $\sqrt[3]x = x^{\frac{1}{3}}$. With this, we can now apply the power rule as shown below.

\begin{aligned}\dfrac{d}{dx} \left(x^{\frac{1}{3}} \right ) &= \dfrac{1}{3} x^{{\frac{1}{3} -1 }} \\&= \dfrac{1}{3} x^{{-\frac{2}{3}}}\\&= \dfrac{1}{3} \dfrac{1}{x^{\frac{2}{3}}}\\&= \dfrac{1}{3x^{\frac{2}{3}}} \end{aligned}

Recall that $b^{\frac{m}{n}} = \sqrt[n]{b^m}$, so $\dfrac{d}{dx} \sqrt[3]{x}$ is also equal to $\dfrac{1}{3\sqrt[3]{x^2}}$.

Practice Questions

1. Evaluate the following expressions:
a. $ \dfrac{d}{dx} \left(x^{\displaystyle{8}} \right )$
b. $ \dfrac{d}{dx} \left(x^{\displaystyle{20}} \right )$
c. . $ \dfrac{d}{dx} \left(x^{\displaystyle{-4}} \right )$
2. Evaluate the following expressions:
a. $ \dfrac{d}{dx} \left(8 x^{\displaystyle{5}} \right )$
b. $ \dfrac{d}{dx} \left(\dfrac{1}{3} x^{\displaystyle{12}} \right )$
c.  $ \dfrac{d}{dx} \left(8 x^{\displaystyle{-8}} \right )$
3. Evaluate the following expressions:
a. $ \dfrac{d}{dx} \left(2e x^{\displaystyle{e}} \right )$
b. $ \dfrac{d}{dx} \sqrt[4]{x}$
c. . $ \dfrac{d}{dx} -4 \sqrt[5]{x^3}$

Answer Key

1.
a. $ 8x^{\displaystyle{7}}$
b. $ 20x^{\displaystyle{19}}$
c. $ – \dfrac{4}{x^{\displaystyle{5}}}$
2.
a. $ 40x^{\displaystyle{4}}$
b. $ 4x^{\displaystyle{11}}$
c. $ – \dfrac{64}{x^{\displaystyle{9}}}$
3.
a. $2e^2 x^{\displaystyle{e – 1}}$
b. $\dfrac{1}{4x^{\frac{3}{4}}} = \dfrac{1}{4\sqrt[4]{x^3}}$
c. $- \dfrac{12 x^{ {12}}}{5x^{\frac{12}{5}}}$