The Derivative as a Limit

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Lesson Priority: High

  • Know what the derivative of a function is and what it means
  • Define the derivative using the difference quotient
  • Use the difference quotient and limits to determine a function's instantaneous slope at a point
Lesson Description

Using what we know about the slope formula, AROC, and limits, we are ready to define a function's exact instantaneous slope at a point using a limit.

Practice Problems

Practice problems and worksheet coming soon!


What is the derivative?

When we looked at The Slope Problem », we said that in first semester Calculus, the problem we are trying to solve is quantifying the slopes of non-linear functions, despite the fact that they constantly vary. When we talk about the derivative of a function, we are referring to finding the tangent slope at a specific point.Above is the graph of $f(x)=x^2$. In this example, we can see (by eyeballing it, for now) that the red line, which is the tangent line at $x=1$, looks to have a slope of $2$, while the orange line, which is the tangent line at $x=2$, looks to have a slope of $4$. Learning what derivatives are and how they work allows us to find these results formulaically, and without requiring graphing or estimation.

Defining the Derivative

Recently, we looked at the average rate of change of a function over a specified interval (AROC »). The definition of the AROC was no more complicated than any slope calculation between two points that we’ve seen in the past. Here’s a reminding example: $$\mathrm{AROC} = \frac{f(x_2)-f(x_1)}{x_2-x_1}$$(1)The dashed red line represents the AROC between $x_1$ and $x_2$.What would happen if $x_1$ and $x_2$ were really, really close together?The dashed gold line represents the tangent line at $x_1$, and the dashed purple line represents the AROC between $x_1$ and $x_2$. Notice how since $x_2$ moved closer to $x_1$, the average rate of change between $x_1$ and $x_2$ is even closer to the tangent line than the red dashed line AROC in the prior figure. Through the magic of limits, we can actually obtain the tangent slope at $x_2$ by calculating the AROC with the limit as $x_1$ becomes infinitely close to $x_2$: $$\lim_{x_1 \to x_2} \frac{f(x_2)-f(x_1)}{x_2-x_1}$$(2)This is one of several similar ways to define the derivative. Limit evaluation techniques are required – it’s not as simple as plugging in $x_2$ in place of $x_1$. If we did, we would have$$\frac{f(x_2)-f(x_2)}{x_2-x_2} = \frac{0}{0}$$As we saw in the recent lesson about limits with zero in the denominator », $0 \div 0$ doesn’t tell us anything, because $0 \div 0$ could be anything. However, we will be able to compute this derivative limit, as long as we specifically know what the function is.

The Classic Difference Quotient

Because it is potentially confusing to work with two different $x$ values ($x_1$ and $x_2$), the limit definition of the derivative is typically defined at a single point on the function where $x=a$, and the secant line (AROC) is measured between $a$ and $a+h$, where $h$ is a small amount of distance from $a$.Therefore, again using the classic slope formula, we can represent the secant slope between $a$ and $a+h$ as$$\frac{f(a+h)-f(a)}{(a+h) - a} = \frac{f(a+h)-f(a)}{h}$$The instantaneous tangent slope at $x=a$ would be obtained if we let $h$ become close to $0$, so that $a+h$ becomes infinitely close to $a$.$$\boxed{\lim_{h \to 0} \frac{f(a+h)-f(a)}{h}}$$This result is the payoff: it is commonly referred to as the Difference Quotient, and when we evaluate this limit for any given function at a specific $x$ value, the result we get is the exact, instantaneous tangent slope of the function at that $x$ value.
Pro Tip
Make sure you can see the connection between the classic slope formula$$m = \frac{y_2 - y_1}{x_2 - x_1}$$and the Difference Quotient. Typically you are required to memorize the Difference Quotient, which is a much easier task if you can think of it as a "rise over run" slope formula. Try to focus on where this formula comes from to aide your memory, instead of just focusing on memorizing symbols.
As an example, let’s revisit our classic parabola $f(x)=x^2$ that we looked at above in the first graph of this lesson.
Example 1Find the tangent slope of $f(x) = x^2$ at $x=2$. $\blacktriangleright$ By eyeballing the graph, we suspected that if we were to draw a tangent line at $x=2$, that the slope of that line would be $4$. Using the Difference Quotient formulaic approach, we can know for sure.The Difference Quotient formula requires is to know $f(a+h)$ and $f(a)$, and $a$, the specific value at which we want to know the slope, is $2$ in this case. Let’s go ahead and compute $f(2+h)$ and $f(2)$, given that $f(x) = x^2$: $$f(2+h) = (2+h)^2 = (2+h)(2+h) = 4 + 4h + h^2$$$$f(2) = (2)^2 = 4$$Finally, we plug these results into the Difference Quotient limit derivative formula.$$\begin{array}{l} \lim_{h \to 0} \frac{f(a+h)-f(a)}{h} \\ =\lim_{h\to 0} \frac{(4 + 4h + h^2)-(4)}{h} \\ =\lim_{h \to 0} \frac{4h + h^2}{h} = \lim_{h \to 0} \frac{h(4 + h)}{h} \\ =\lim_{h \to 0} 4 + h = \boxed{4} \end{array} $$The instantaneous slope of $f(x) = x^2$ at $x=2$ is $4$, as we supposed.Notice how the $h$ ended up cancelling – this will always happen when you’re asked to find the derivative this way. Notice also that every step along the way, the limit notation remains in front of the expression, until the step where we can actually take the limit (after the $h$ canceled). Not only must you keep detailed scratch work, but you MUST include this limit step along the way until you take the limit! Teachers love to dock points for neglecting it, and losing points for it is needless and will make you sad.
You Should Know:
Teachers use this technique as a way of introducing the derivative as a concept, and you will very likely need to be able to compute slopes and general derivatives using the Difference Quotient limit formula. However, you will soon be learning severely amazing shortcuts for finding derivatives, so don’t become too attached to this method.

Put It To The Test

One way that the limit definition of the derivative is commonly tested is to be given polynomial, rational, or root functions, and be asked to find the derivative using the limit definition. One thing that's nice about these problems is that they are all very similar - that is, each category has its own process and all problems in that category are done identically. These forms will look a little familiar from work we did in the lesson on limits with zero in the denominator ». Let's see one of each.
Example 4Using the limit definition of the derivative, find the slope of the following function at the point where $x = -2$.$$f(x) = 5-2x-3x^2$$
Show solution
$\blacktriangleright$ Given the definition of $f(x)$, we need to figure out what $f(a)$ and $f(a+h)$ each is. For this problem, our $a$ is $-2$, since that's the $x$ value at which I want to know the tangent slope.$$\lim_{h \to 0} \frac{f(a+h)-f(a)}{h}$$$$=\lim_{h \to 0} \frac{f(-2+h)-f(-2)}{h}$$Calculate these components separately, then plug them into this formula.$$f(-2) = 5 - 2(-2) - 3(-2)^2 = \boxed{-3}$$$$f(-2+h) = 5 - 2(-2+h) - 3(-2+h)^2$$$$=5+4-2h - 3(4-4h+h^2) = \boxed{-3+10h-3h^2}$$Let's plug in these results and manipulate the expression so that we can get the original $h$ in the denominator to cancel.$$\lim_{h \to 0} \frac{\left(-3+10h-3h^2\right) - (-3)}{h}$$$$=\lim_{h \to 0} \frac{10h - 3h^2}{h}$$$$=\lim_{h \to 0} \frac{h(10 - 3h)}{h}$$$$=\lim_{h \to 0} \frac{\cancel{h}(10 - 3h)}{\cancel{h}}$$$$=\lim_{h \to 0} 10 - 3h = \boxed{10}$$Therefore, the slope of $f(x) = 5-2x-3x^2$ at $x=-2$ is $10$.
Pro Tip
All polynomial functions are done this way, when finding the derivative via the limit definition. In other words, when doing this for a polynomial you will always plug in the appropriate expressions and find that all the terms in the numerator without an $h$ will cancel with one another, leaving you to finish the problem by factoring out $h$ in the numerator, and cancelling it with the $h$ in the denominator.
Example 5 Using the limit definition of the derivative, find the slope of the following function at the point where $x = 2$.$$f(x) = \frac{1}{2x-1}$$
Show solution
$\blacktriangleright$ Just like with the previous polynomial example, we need to work with the Difference Quotient in order to follow our explicit instructions to use the "limit definition of the derivative". The only difference (no pun intended) that we'll see with this example versus the previous one is that the $h$ will not cancel so easily - we'll have to do some algebra manipulation to make it happen.$$f(2) = \frac{1}{2(2)-1} = \frac{1}{3}$$$$f(2 + h) = \frac{1}{2(2+h)-1} = \frac{1}{3+2h}$$$$\lim_{h \to 0} \frac{f(a+h)-f(a)}{h}$$$$=\lim_{h \to 0} \frac{f(2+h)-f(2)}{h}$$$$=\lim_{h \to 0} \frac{\frac{1}{3+2h} - \frac{1}{3}}{h}$$This is quickly starting to look like a mess because we have fractions within fractions. This initial complex fraction result is unavoidable for rational functions, but we can always choose to do what I am going to recommend here: take out the $h$ denominator as a fraction, and rewrite this expression as a product of fractions, instead of a mess of fractions within fractions.$$\lim_{h \to 0} \frac{f(a+h)-f(a)}{h} = \lim_{h \to 0} \frac{1}{h} \,\, \left[ f(a+h)-f(a)\right]$$$$\Rightarrow \lim_{h \to 0} \frac{1}{h} \,\, \left[ \frac{1}{3+2h} - \frac{1}{3} \right]$$Again, we can always do this with rational function Difference Quotient calculations. Now, we'll proceed by doing what we often need to do when adding or subtracting fractions: obtain identical denominators.$$\lim_{h \to 0} \frac{1}{h} \,\, \left[ \frac{1}{3+2h} \cdot \frac{3}{3} - \frac{1}{3} \cdot \frac{3+2h}{3+2h} \right]$$$$\lim_{h \to 0} \frac{1}{h} \,\, \left[ \frac{3}{3(3+2h)} - \frac{3 + 2h}{3(3+2h)} \right]$$$$\lim_{h \to 0} \frac{1}{h} \,\, \left[ \frac{3-3-2h}{3(3+2h)} \right]$$$$\lim_{h \to 0} \frac{1}{\cancel{h}} \,\, \left[ \frac{-2\cancel{h}}{3(3+2h)} \right]$$$$\lim_{h \to 0} \frac{-2}{3(3+2h)} = \frac{-2}{3(3)}$$$$\boxed{-\frac{2}{9}}$$
Pro Tip
Every rational function you should expect to see on a test is handled this same way. And for your own sanity, I highly recommend you adopt my approach of keeping the original $h$ denominator as a factored out $1/h$ term, so that you can work with the rest of the expression without stressing about fractions within fractions. Obtain a single fraction with a common denominator, like I did with this problem, and you'll be on your way in no time.
Example 6 Using the limit definition of the derivative, find the slope of the following function at the point where $x = 4$.$$f(x) = \sqrt{x+3}$$
Show solution
$\blacktriangleright$ As we've seen, we need to evaluate the function at $4$ and $4+h$, in order to plug those results into the Difference Quotient.$$f(4) = \sqrt{4+3} = \sqrt{7}$$$$f(4 + h) = \sqrt{4 + h + 3} = \sqrt{7 + h}$$As before, we'll plug these into the Difference Quotient.$$\begin{array}{l} \lim_{h \to 0} \frac{f(a+h)-f(a)}{h} \\ =\lim_{h \to 0} \frac{f(4+h)-f(4)}{h} \\ =\lim_{h \to 0} \frac{\sqrt{7 + h} - \sqrt{7}}{h} \end{array} $$Just like we saw in the previous example with a rational function, here we will not instantly have a way to make that denominator $h$ cancel. Instead, we must play a similar game to rational functions: algebra manipulation. Here we will use the property of radical conjugates, and multiply this fraction top and bottom by $\sqrt{7 + h} + \sqrt{7}$, which is the conjugate of $\sqrt{7 + h} - \sqrt{7}$.$$\lim_{h \to 0} \frac{\sqrt{7 + h} - \sqrt{7}}{h} \cdot \frac{\sqrt{7 + h} + \sqrt{7}}{\sqrt{7 + h} + \sqrt{7}}$$$$=\lim_{h \to 0} \frac{(7+h) +\sqrt{7}\sqrt{7+h} - \sqrt{7}\sqrt{7+h} - (7)}{h \left[ \sqrt{7 + h} + \sqrt{7} \right]}$$$$=\lim_{h \to 0} \frac{(7+h) - (7)}{h \left[ \sqrt{7 + h} + \sqrt{7} \right]}$$$$=\lim_{h \to 0} \frac{\cancel{h}}{\cancel{h} \left[ \sqrt{7 + h} + \sqrt{7} \right]}$$$$=\lim_{h \to 0} \frac{1}{\sqrt{7 + h} + \sqrt{7}}$$$$= \frac{1}{\sqrt{7} + \sqrt{7}} = \boxed{\frac{1}{2\sqrt{7}}}$$Skip steps at your own peril - many teachers will dock points for doing too many of these steps at once.
Pro Tip
Every root function a teacher will put on an exam (very, very very likely anyway) is done identically to how this problem was done. In short, you always multiply the whole expression both top and bottom by the radical conjugate of the numerator. The numerator will be left with no roots. The denominator will look a bit sad at first, but cleans up nicely when you get to the step where you can allow $h$ to be zero.
Lesson Takeaways
  • Know what the Difference Quotient is and where it comes from
  • Be able to evaluate the Difference Quotient limit at a specific value, and know what it means
  • Don’t lose points by forgetting the lim in your scratch work! Don’t be lazy!
  • Practice doing this for polynomial, rational, and radical functions, as all three are feasible test questions

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