5.3 Derivative of Polynomial Functions

There are a few rules to remember to find the derivative of a function. Before we hop into those rules, we will start by finding the derivative of some basic functions.

Suppose we have the function

$f(x)=c$

where $c$ is a constant, real number. Based on the definition of a derivative,

$f^{\prime }(x)=\underset{h\to 0}{lim}\frac{f(x+h)-f(x)}{h}=\underset{h\to 0}{lim}\frac{c-c}{h}=0$

This shows us that the derivative of any constant is 0.

Now let’s consider three functions:

$f(x)=x^2$

$f(x)=\sqrt{x}$

$f(x)=\frac{1}{x}$

Let’s use the definition of a derivative to find the derivative of each function.

Here’s the first function:

$\begin{array}{rl}{f}^{\prime }(x)& =\underset{h\to 0}{lim}\frac{f(x+h)-f(x)}{h}\\ & =\underset{h\to 0}{lim}\frac{(x+h{)}^2-x^2}{h}\\ & =\underset{h\to 0}{lim}\frac{x^2+2xh+h^2-x^2}{h}\\ & =\underset{h\to 0}{lim}\frac{\frac{x^2+2xh+h^2-x^2}{h}}{\frac{h}{h}}\\ & =\underset{h\to 0}{lim}(2x+h)\\ & =2x\end{array}$

From this example, we can see that the derivative of a second-degree polynomial is found by decreasing the power of $x$ by 1 and then multiplying $x$ by the original exponent.

How about roots? Let’s look at our second example.

$\begin{array}{rl}{f}^{\prime }(x)& =\underset{h\to 0}{lim}\frac{f(x+h)-f(x)}{h}\\ & =\underset{h\to 0}{lim}\frac{\sqrt{x+h}-\sqrt{x}}{h}\\ & =\underset{h\to 0}{lim}\frac{(\sqrt{x+h}-\sqrt{x})(\sqrt{x+h}+\sqrt{x})}{h(\sqrt{x+h}+\sqrt{x})}\\ & =\underset{h\to 0}{lim}\frac{x+h-x}{h(\sqrt{x+h}+\sqrt{x})}\\ & =\underset{h\to 0}{lim}\frac{h}{h(\sqrt{x+h}+\sqrt{x})}\\ & =\underset{h\to 0}{lim}\frac{1}{(\sqrt{x+h}+\sqrt{x})}\\ & =\frac{1}{(\sqrt{x}+\sqrt{x})}\\ & =\frac{1}{2\sqrt{x}}\\ & =\frac{1}{2}{x}^{-\frac{1}{2}}\end{array}$

If we remember that $f(x)=\sqrt{x}=x^{\frac{1}{2}}$, we can see a pattern similar to the derivative of $f(x)=x^2$.

Finally, let’s look at a rational function. Here’s our third example:

$\begin{array}{rl}{f}^{\prime }(x)& =\underset{h\to 0}{lim}\frac{f(x+h)-f(x)}{h}\\ & =\underset{h\to 0}{lim}\frac{\frac{1}{x+h}-\frac{1}{x}}{h}\\ & =\underset{h\to 0}{lim}\frac{\frac{x}{(x+h)x}-\frac{x+h}{(x+h)x}}{h}\\ & =\underset{h\to 0}{lim}\frac{\frac{-h}{(x+h)x}}{h}\\ & =\underset{h\to 0}{lim}\frac{-1}{(x+h)x}\\ & =-\frac{1}{x^2}\\ & =-x^{-2}\end{array}$

One thing to remember for the second function is that $f(x)=\frac{1}{x}=x^{-1}$. Can you see the pattern? We will now introduce the first rule of the derivative, which is the power rule.

The Power Rule

If $a$ is any real number, then the derivative of the function $f(x)=x^a$ is

$f^{\prime }(x)=a{x}^{a-1}$

If $a=1$, or if $f(x)=x$, then the derivative of it is

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

Let’s look at some examples. What is the derivative of $f(x)=x^{10}$? As we follow the power rule,

$f^{\prime }(x)=10x^9$

How about the derivatives of $f(x)=\sqrt[4]{x^3}$ and $g(x)=\frac{1}{x^4}$ ? Remember that $f(x)=\sqrt[4]{x^3}=x^{\frac{3}{4}}$ and $g(x)=\frac{1}{x^4}=x^{-4}$. As we follow the power rule,

$\begin{array}{rl}{f}^{\prime }(x)& =\frac{3}{4}{x}^{\frac{3}{4}-1}\\ & =\frac{3}{4}{x}^{-\frac{1}{4}}\end{array}$

$\begin{array}{rl}{g}^{\prime }(x)& =-4x^{-4-1}\\ & =-4x^{-5}\end{array}$

We are going to introduce two more rules: the derivative of a function multiplied by a constant, and the derivative of a sum or difference of two functions.

Derivative of a Function Multiplied by a Constant

If a function is multiplied by a constant $c$, then the derivative of $cf(x)$ is

$\frac{d}{dx}[cf(x)]=c\frac{d}{dx}[f(x)]$

Derivative of the Sum or Difference of Two Functions

If we are taking a derivative of $f(x)+g(x)$ or $f(x)-g(x)$ with respect to $x$,

$\frac{d}{dx}[f(x)\pm g(x)]=\frac{d}{dx}[f(x)]\pm \frac{d}{dx}[g(x)]$

Practice

Let’s practice finding the derivative of polynomial functions. Can you find the derivatives of the following two functions?

$f(x)=3x^4-2x^2+x-1$

$g(x)=\frac{5}{x^5}$

Here is the solution for the first function.

$\begin{array}{rl}{f}^{\prime }(x)& =3x^4-2x^2+x-1\\ & =4\cdot 3x^{4-1}-2\cdot 2x^{2-1}+1\cdot x^{1-1}-0\\ & =12x^3-4x+1\end{array}$

For the second function, since

$\begin{array}{rl}g(x)& =\frac{5}{x^5}\\ & =5x^{-5}\end{array}$

the derivative with respect to $x$ is

$\begin{array}{rl}{g}^{\prime }(x)& =-5\cdot 5x^{-5-1}\\ & =-25x^{-6}\\ & =\frac{-25}{x^6}\end{array}$

Derivative of an Exponential Function

Now, let’s look at the derivative of an exponential function. Since this is not meant to be a course for those who are going to study advanced mathematics, we will give you the following definition of the number $e$:

$\underset{h\to 0}{lim}\frac{e^h-1}{h}=1$

Given the above definition, we can find the derivative of a function $f(x)=e^x$ with respect to $x$.

$\begin{array}{rl}{f}^{\prime }(x)& =\underset{h\to 0}{lim}\frac{f(x+h)-f(x)}{h}\\ & =\underset{h\to 0}{lim}\frac{e^{x+h}-e^x}{h}\\ & =\underset{h\to 0}{lim}\frac{e^x{e}^h-e^x}{h}\\ & =\underset{h\to 0}{lim}\frac{e^x(e^h-1)}{h}\\ & =e^x\underset{h\to 0}{lim}\frac{e^h-1}{h}\\ & =e^x\end{array}$

In a later section of this topic (when we learn about the chain rule), we will discuss the derivative of the function $f(x)=e^{g(x)}$.

Derivative of a Natural Log

Finally, what is the derivative of $f(x)=\ln x$?

$\begin{array}{rl}{f}^{\prime }(x)& =\underset{h\to 0}{lim}\frac{f(x+h)-f(x)}{h}\\ & =\underset{h\to 0}{lim}\frac{\ln (x+h)-\ln x}{h}\\ & =\underset{h\to 0}{lim}\frac{\ln \frac{x+h}{x}}{h}(\text{Recall}\ln a-\ln b=\ln \frac{a}{b})\\ & =\underset{h\to 0}{lim}\frac{\ln (1+\frac{h}{x})}{h}\\ & =\underset{h\to 0}{lim}\frac{1}{h}\ln (1+\frac{h}{x})\\ & =\underset{h\to 0}{lim}\ln {(1+\frac{h}{x})}^{\frac{1}{h}}(\text{Recall}c\ln a=\ln a^c)\end{array}$

Now let $n=\frac{h}{x}$ . So $\frac{1}{h}=\frac{1}{nx}$ , and as $h$ approaches 0, $n$ also approaches 0. Therefore, we have

$\begin{array}{rl}{f}^{\prime }(x)& =\underset{n\to 0}{lim}\ln {(1+n)}^{\frac{1}{n}\cdot \frac{1}{x}}\\ & =\underset{n\to 0}{lim}\frac{1}{x}\ln {(1+n)}^{\frac{1}{n}}\\ & =\frac{1}{x}\underset{n\to 0}{lim}\ln {(1+n)}^{\frac{1}{n}}(\text{Recall}\underset{n\to 0}{lim}\ln {(1+n)}^{\frac{1}{n}}=e)\\ & =\frac{1}{x}\ln e\\ & =\frac{1}{x}\end{array}$

Given this result, the derivative of $f(x)={\log }_{b}x$ is

$f^{\prime }(x)=\frac{1}{x\ln b}$

This is because

$\begin{array}{rl}f(x)& ={\text{log}}_{b}x\\ & =\frac{\ln x}{\ln b}(\text{from the change of base formula})\\ & =\frac{1}{\ln b}\ln x\end{array}$

where $\frac{1}{\ln b}$  is a constant and the derivative of $\ln x$ is $\frac{1}{x}$ .