Power functions are absolutely everywhere! You will find these in the vast majority of problems you will encounter in business calculus. This means that you will be using it very often, and it will quickly become second nature to you after some practice.

As explained above, the Power Rule is used to find derivatives of power functions, which are functions that look like $x^n$, where $n$ is a number. So, for example, the following are power functions:

$$f(x) = x^2$$

$$f(x) = x^5$$

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

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

$$f(x) = x^{5.5}$$

As you can see, the exponent of a power function can be positive or negative integer, or even a fraction or decimal too.

Now, in the examples above, we only have the variable $x$ raised to an exponent. But what happens if we multiply the power function by a coefficient (a number)? Then we get a generalized power function, which is in the form $f(x)=ax^n$, where $a$ is a number.

Here are some examples of generalized power functions:

$$f(x) = 2x^7$$

$$f(x) = 3x^2$$

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

In other words, generalized power functions are power functions with a number multiplying in the front. That number is called the coefficient. For example, for the function $f(x)=2x^3$, the coefficient is $2$, and the power (the exponent) is $3$.

The Power Rule can be used for both power functions and generalized power functions. Actually, if you think about it, a power function is a special case of a generalized power function with a coefficient equal to $1$. For example, the following is a generalized power function:

$$f(x)=1 \cdot x^2$$

But if we simplify by multiplying by $1$, it just becomes a power function since something times $1$ is equal to the same thing:

$$f(x)=1 \cdot x^2 = x^2$$

Based on the above, you have to use the Power Rule whenever you see a function in the form of $f(x)=ax^n$, where $a$ and $n$ are numbers.

Below is an explanation on how to differentiate power functions.

In words: The derivative of a generalized power function, $f(x) = a \cdot x^n$ (where $a$ and $n$ are numbers), is equal to bringing down to the front the exponent $n$ multiplying by the coefficient $a$, and subtracting $1$ from the exponent

In math notation: If $f(x) = ax^n$, then $f'(x)=n \cdot ax^{(n-1)}$

Let's examine a sample problem below.

Given that $f(x)=3x^5$,

Question 1Find $f'(x)$

We know this is a generalized power function with coefficient $3$ and exponent $5$. According to the Power Rule, we have to bring down the exponent multiplying, and then subtract one from the exponent,

$$f'(x)=\color{red}{5} \cdot 3x^{(\color{red}{5-1})}$$

See that red $5$ multiplying everything? That is the original exponent, which came down multiplying. And see that $(5-1)$ exponent? That is the original exponent minus $1$. We just applied the Power Rule! Now we simplify.

$$f'(x)=5 \cdot 3x^{(5-1)}=15x^4$$

(in case you missed it, we multiplied the coefficients $5 \cdot 3 = 15$, and subtracted $1$ from the exponent, leaving $5-1=4$ as the new exponent)

There is one special case that we need to discuss to avoid confusion, which is when $x$ is elevated to the first power. In other words, when for example we have $f(x)=4x$. In this case, it looks like $x$ has no exponent. But when you see a letter without an exponent, it just means that its exponent is $1$. So, $f(x)=4x=4x^1$

To get the derivative of $f(x)=4x$, we apply the usual process: bring down the exponent multiplying, and subtract $1$ from the exponent,

$$f'(x)=\color{red}{1} \cdot 4x^{(\color{red}{1-1})}=4x^0$$

So now we have $x$ to the zero. What does that equal to? Remember that anything elevated to the zero power equals $1$. So then,

$$f'(x)=4x^0=4 \cdot (1) = 4$$

- The Power Rule is used to find the derivatives of generalized power functions
- A generalized power function is in the form $f(x)=a \cdot x^n$, where $a$ and $n$ are numbers
- A generalized power function whose coefficient $a$ equals $1$ is called just a power function, and it's in the form $f(x)=x^n$, where $n$ is a number
- To apply the Power Rule, bring down the exponent multiplying the coefficient, and then subtract $1$ from the current exponent