The Derivative as a Function

Lesson Features »

Lesson Priority: High

  • Obtain a function's "slope function" using derivatives
  • Use the difference quotient to get the "slope function" of a function for basic polynomial, radical, and rational functions
  • Understand why the generic slope function is better than finding specific slopes
Lesson Description

The limit approach to instantaneous slope from the last lesson got us exact slopes at a specific points. This lesson instead will show us how to obtain a function's "slope function" which will yield us the slope of the function at any point without having to repeatedly use the limit definition.

Practice Problems

Practice problems and worksheet coming soon!


Generalizing Derivatives

In the lesson immediately prior », we looked at what derivatives are at specific locations on a function using the limit definition of the derivative (essentially, they are instantaneous slopes). It makes sense to start with looking at derivatives at a specific place on a function - after all, if we are interested in measuring a function's particular instantaneous slope, then we need to do it at a particular location. Next up, we'll see how to breed generic results that can be used over and over again for each function. Instead of doing a calculation at a specific $x$ value, we can generalize the difference quotient process to create a new function - the derivative function.Before we understand how to obtain a function's derivative function, let's better define it.
Define: The Derivative FunctionLet $f(x)$ be a continuous and smooth function. There exists another function notated $f'(x)$ (pronounced "f prime of x"), which is related to $f(x)$ in that $f'(x)$ describes the instantaneous slope of $f(x)$ at a given $x$ value, similarly to how $f(x)$ describes the function value at a given $x$ value.We say that $f'(x)$ is the derivative function to $f(x)$. Furthermore, each such function $f(x)$ has one definitive $f'(x)$. That is, there are not multiple possible derivative functions for a given function - there is only one.
In other words, if you know $f(x)$, there will be some process we can follow to figure out what its correct $f'(x)$ is. Note that in most cases, $f'(x)$ is the same general function category as $f(x)$, but not typically equal to $f(x)$ (in fact there is only one function for which $f(x) = f'(x)$).Let's see a quick example of this concept for $f(x) = x^2$, without explaining for now how I know that $f'(x) = 2x$ for $f(x) = x^2$ (it is - and you can prove it to yourself using the process we'll see later in this lesson).
Example 1Describe the relationship between $f(x)$ and $f'(x)$ for the given function at $x=3$.$$\blacktriangleright \,\, \begin{array}{rl} f(x) & = x^2 \\ f'(x) & = 2x \end{array}$$From what we know about functions, when we evaluate $f(x)$ at $x = 3$ to obtain$$f(3) = 3^2 = 9$$we know that we are saying that the function value is $9$ when $x$ is $3$. Now, with the introduction of the derivative function $f'(x)$, we say that since $f'(x) = 2x$, that $f'(3) = 2(3) = 6$, and the meaning is that when $x$ is $3$, the slope of $f(x)$ is $6$. Let's look at a graph for reinforcement.As we can see, the function value is $9$ when $x$ is $3$. The red dashed line shows the tangent slope at $x=3$, and if we count units to estimate rise over run, it does indeed appear that the slope of that line is $6$.
You Should Know
Right now we are focusing on understanding that a given function has a related derivative function, and that a function's derivative function gives slopes of the original function, just the same way that the original function gives values.We are not yet focusing on how to figure out what a function's derivative is yet. Several of the next upcoming lessons will teach us how to actually find derivatives based on what kind of function you are working with.Polynomials »Trig Functions »Exponentials »Logarithms »Mastering Derivatives of Any Function »

Obtaining The Derivative

The limit definition of the derivative at a specific point (the topic of the immediately prior lesson ») works well for deriving the instantaneous slope of a function at a specific point. However, this would be a very inefficient process if we wanted to measure several instantaneous slopes of the same function, because there is a way to find a general expression for a function's derivative using the difference quotient, and it looks incredibly similar to the work we did at a specific $x$ value.
Define: The "Limit Definition" of the Derivative FunctionLet $f(x)$ be a continuous and smooth function for all $x$. It follows that$$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$where $f'(x)$ is the derivative function for the function $f(x)$.
Let's use the difference quotient to find the derivative of a function using the limit definition, similar to what we did in the last lesson. The only difference is, instead of working with the difference quotient at a specific $x$ value, we will leave the function as-is, without evaluating it. The result we get will be the derivative.
Example 2Find the derivative function of $f(x) = 3x^2 - 4x + 1$ using the limit definition of the derivative. Then find the slope of this function at $x=-3$, $x = 0$, and $x = 5$.$\blacktriangleright$ Recall what the difference quotient looks like:$$\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$Unlike the last lesson, we will not pick $x$ to be a specific value. Instead, we will plug $f(x)$ and $f(x+h)$ directly into the limit, and simplify down until we can get an $h$ to cancel top and bottom - this will be the part that looks incredibly similar to our work in the last lesson.$$\begin{array}{rl} f(x) & = 3x^2 - 4x + 1 \\ f(x + h) & = 3(x+h)^2 - 4(x + h) + 1 \\ & = 3\left(x^2 + 2xh + h^2 \right) - 4x - 4h + 1 \\ & = 3x^2 + 6xh + 3h^2 - 4x - 4h + 1 \end{array}$$Now let's plug in.$$\begin{array}{rl} f'(x) & = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \\ & = \lim_{h \to 0} \frac{\left(3x^2 + 6xh + 3h^2 - 4x - 4h + 1\right) - \left( 3x^2 - 4x + 1 \right)}{h} \\ & = \lim_{h \to 0} \frac{6xh + 3h^2 - 4h}{h} \\ & = \lim_{h \to 0} \frac{\cancel{h} \big( 6x + 3h - 4\big)}{\cancel{h}} \\ & = \lim_{h \to 0} 6x + 3h - 4 \\ & = 6x - 4 \end{array}$$Therefore, for $f(x) = 3x^2 - 4x + 1$, we have found that $f'(x) = 6x - 4$.To finish the problem, let's use our newly found derivative function to find the instantaneous slope of $f(x)$ at the specified points.At $x=-3$:$$\begin{array}{rl} f'(-3) & = 6(-3) - 4 \\ & = -18 - 4 = -22 \end{array}$$Therefore the slope of $f(x)$ at $x = -3$ is $-22$.At $x=0$:$$\begin{array}{rl} f'(0) & = 6(0) - 4 \\ & = - 4 \end{array}$$Therefore the slope of $f(x)$ at $x = 0$ is $-4$.At $x=5$:$$\begin{array}{rl} f'(5) & = 6(5) - 4 \\ & = 30 - 4 = 26 \end{array}$$Therefore the slope of $f(x)$ at $x = 5$ is $26$.
Pro Tip
This work is incredibly similar to the work we did in the last lesson ». The only difference is that in the last lesson, we had a specific $x$ value in mind at which to examine the tangent slope, while here, we are performing the limit in order to obtain a variable expression result. The work is a little less pleasant to look at since we will have two variables (both $x$ and $h$) in our calculations throughout our scratch work, while we only had $h$ to deal with in the last lesson (since we had already picked a static numerical value of $x$).An important takeaway is that if we are asked to do a "limit definition of the derivative" in any way, we will ultimately be setting up and evaluating a limit of $h$ going to $0$ of some kind of difference quotient, whether it's at a specific location as we did in the last lesson, or obtaining the generic derivative function as we are doing now.

Put It To The Test

Similar to what we saw in the last lesson, you should be prepared to derive polynomial, radical, and rational functions' derivatives using the limit definition. Let's practice one of each.
Example 3For $f(x) = x^3 - 2x^2 + 5$, find $f'(x)$.
Show solution
$\blacktriangleright$ To obtain the derivative function $f'(x)$ using the limit definition of the derivative (note that you could use shortcuts if you know them, since the instructions didn't specify to use the limit definition), we must calculate $f(x+h)$ and correctly plug the pieces into the difference quotient.$$\begin{array}{rl} f(x) & = x^3 - 2x^2 + 5 \\ f(x+h) & = (x+h)^3 - 2(x+h)^2 + 5 \end{array} $$Every time we work with a polynomial for this process, it is important to expand the binomials. You won't get anywhere if you leave $(x+h)^3$ as $(x+h)^3$.$$\begin{array}{rl} f(x+h) & = (x^3 + 3x^2 h + 3x h^2 + h^3) -2(x^2 + 2xh + h^2) + 5 \\ & = x^3 + 3x^2 h + 3x h^2 + h^3 - 2x^2 - 4xh - 2h^2 + 5 \end{array} $$Now, plug in.$$\begin{array}{rl} f'(x) & = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \\ & =\lim_{h \to 0} \frac{\big(x^3 + 3x^2 h + 3x h^2 + h^3 - 2x^2 - 4xh - 2h^2 + 5\big) - \big(x^3 - 2x^2 + 5\big)}{h} \\ & =\lim_{h \to 0} \frac{3x^2 h + 3x h^2 + h^3 - 4xh - 2h^2}{h} \\ & =\lim_{h \to 0} \frac{h \big(3x^2 + 3xh + h^2 -4x -2h\big)}{h} \\ & =\lim_{h \to 0} \frac{\cancel{h} \big(3x^2 + 3xh + h^2 -4x -2h\big)}{\cancel{h}} \\ & =\lim_{h \to 0} 3x^2 + 3xh + h^2 -4x -2h \\ & = 3x^2 - 4x \end{array} $$Therefore, the derivative function for $f(x) = x^3 - 2x^2 + 5$ is $f'(x) = 3x^2 - 4x$.
Example 4For $f(x) = \sqrt{2x+7}$ find $f'(x)$ using the limit definition.
Show solution
$\blacktriangleright$ We'll need to start with the same steps.$$\begin{array}{rl} f(x+h) & = \sqrt{2(x+h) + 7} \\ & = \sqrt{2x + 2h + 7} \end{array} $$We will once again plug into the difference quotient, and when we're done simplifying fully, we'll have the derivative function.$$\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$$$\lim_{h \to 0} \frac{\sqrt{2x + 2h + 7} - \sqrt{2x + 7}}{h}$$Unlike polynomials, there is no simplistic or automatic cancellation of terms that gets us to where we need to be. We will have to employ an algebra manipulation technique based on the "difference of squares" factoring principle. This will look familiar from an example or two in the last lesson.$$\lim_{h \to 0} \frac{\sqrt{2x + 2h + 7} - \sqrt{2x + 7}}{h} \cdot \frac{\sqrt{2x + 2h + 7} + \sqrt{2x + 7}}{\sqrt{2x + 2h + 7} + \sqrt{2x + 7}}$$$$\lim_{h \to 0} \frac{(2x + 2h + 7)-(2x + 7)}{h \big[ \sqrt{2x + 2h + 7} + \sqrt{2x + 7}\big]}$$$$\lim_{h \to 0} \frac{2\cancel{h}}{\cancel{h} \big[ \sqrt{2x + 2h + 7} + \sqrt{2x + 7}\big]}$$$$\lim_{h \to 0} \frac{2}{\sqrt{2x + 2h + 7} + \sqrt{2x + 7}}$$$$=\frac{2}{\sqrt{2x + 7} + \sqrt{2x + 7}}$$$$=\frac{2}{2\sqrt{2x + 7}}$$$$=\frac{1}{\sqrt{2x+7}}$$Therefore, $f'(x) = \frac{1}{\sqrt{2x+7}}$ for $f(x) = \sqrt{2x+7}$.
Example 5For $f(x) = \frac{1}{x^2+1}$ find $f'(x)$ using the limit definition.
Show solution
$\blacktriangleright$ Again, we need $f(x+h)$.$$\begin{array}{rl} f(x+h) & = \frac{1}{(x+h)^2 + 1} \\ & = \frac{1}{x^2 + 2xh + h^2 + 1} \end{array} $$Now we plug into the formula. Keep in mind that with rational functions, it's easier to write the difference quotient as a product than a quotient.$$\lim_{h \to 0} \frac{1}{h} \cdot \big[ f(x+h) - f(x) \big]$$$$\lim_{h \to 0} \frac{1}{h} \cdot \left[ \frac{1}{x^2 + 2xh + h^2 + 1} - \frac{1}{x^2+1} \right]$$We will once again employ particular algebra manipulation tricks to get us where we need to be.$$\lim_{h \to 0} \frac{1}{h} \cdot \left[\frac{1}{x^2 + 2xh + h^2 + 1} \cdot \frac{x^2+1}{x^2 + 1} - \frac{1}{x^2+1} \cdot \frac{x^2 + 2xh + h^2 + 1}{x^2 + 2xh + h^2 + 1} \right]$$$$\lim_{h \to 0} \frac{1}{h} \cdot \frac{(x^2+1) - (x^2 + 2xh + h^2 + 1)}{\big(x^2 + 2xh + h^2 + 1 \big) \big( x^2+1 \big)}$$$$\lim_{h \to 0} \frac{1}{h} \cdot \frac{-2xh - h^2}{\big(x^2 + 2xh + h^2 + 1 \big) \big( x^2+1 \big)}$$$$\lim_{h \to 0} \frac{1}{\cancel{h}} \cdot \frac{\cancel{h}(-2x - h)}{\big(x^2 + 2xh + h^2 + 1 \big) \big( x^2+1 \big)}$$$$\lim_{h \to 0} \frac{(-2x - h)}{\big(x^2 + 2xh + h^2 + 1 \big) \big( x^2+1 \big)}$$$$\frac{(-2x)}{\big(x^2 + 1 \big) \big( x^2+1 \big)}$$$$-\frac{2x}{\left(x^2 + 1\right)^2}$$Therefore, the derivative function $f'(x)$ for the original function $f(x)$ is:$$\begin{array}{ll} f(x) & = \frac{1}{x^2+1} \\ f'(x) & = -\frac{2x}{\left(x^2 + 1\right)^2} \end{array} $$
Lesson Takeaways:
  • Understand the difference between finding the derivative value at a specific point and finding the derivative function
  • Be able to use the limit definition of the derivative to find a given function's generic derivative
  • Similar to prior knowledge, know how to algebraically manipulate polynomials, rational functions, and radical functions with the difference quotient

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