# Differentiability

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Objectives
• Understand what "differentiable" means and how it is similar and different from "continuous"
• See visual cues from graphs that help us define and understand differentiability
• Define differentiability using a limit definition
Lesson Description

Not all functions have well-defined slope functions. This lesson explores the relationship between whether a function has a well-defined slope function and the characteristics of that function.

Practice Problems

Practice problems and worksheet coming soon!

## Smooth AF

While continuity describes a function's "connectedness", differentiability describes a function's "smoothness". Of course, like continuity, we should know the technical definition, since we'll be expected to know and use it.
Define: DifferentiabilityLet $f(x)$ be continuous on the interval $[a,b]$. We say that $f(x)$ is differentiable if$$\lim_{x \to c} f'(x) = f'(c)$$for every point $c$ that is in the interval.
This is incredibly similar to the limit definition of continuity. Recall that $f(x)$ is continuous on $[a,b]$ if$$\lim_{x \to c} f(x) = f(c)$$In fact, another way to say a function is differentiable is to say that its derivative is continuous. Sometimes textbook problems or exams use this as a tricky way to communicate differentiability to us - they'll say "the derivative is continuous" or "$f'(x)$ is continuous on $[a,b]$".
You Should Know Differentiability, a.k.a. function smoothness, is a prerequisite quality for a function to have in order for many Calculus topics to work. If you need a function to be differential in order to apply a rule, make sure you are told that it is. However, problems may state differentiability without using the word differentiability, such as saying "the derivative of $f(x)$ is continuous" or "$f(x)$ has a derivative that is defined everywhere". Also, as you might expect, many common function families are differentiable on their domains.The mean value theorem » is an example of a rule that requires differentiability.

## Looking for Smooth

A graph is an incredibly good tool to help us identify locations of discontinuity. First and foremost, any place where a discontinuity exists will also be a place that is not differentiable. Second, places that are continuous but not differentiable tend to have a certain look to them - they are often corners, cusps, or peaks.Take a look at the following function graph. Since this graph has a jump discontinuity at $x=3$, we know the function is already not continuous at that $x$ value, and therefore it is also not differentiable there either.Other less obvious graphical indicators of places where a function is not differentiable include cusps and corners. Cusps tend to look like a 1st grader's drawing of a seagull: Each side of the cusp has a tangent slope that approaches $\infty$ or $-\infty$.Corners are like cusps but don't need to have vertical tangents. A classic example is the graph of the function $f(x) = |x|$: While the graph does not seem to have any vertical tangency behavior, there is clearly a corner where the graph does not exhibit smoothness, namely at $x=0$.

## What's the "Point"?

For this topic, we will be expected to be able to do a few things - first, conceptually understand differentiability and be able to explain what it is in either words or math symbols. Next, we should be ready to show or demonstrate differentiability using either a graph or algebra approach. Finally, we should understand the concept of differentiability as a yes / no question. It is a "yes" at the place when the function is smooth, and a "no" at the places where the function has discontinuities, corners, or cusps.The common questions for quizzes and tests involve demonstrating whether a function is differentiable by applying either the graph approach or the algebra approach.The Graph ApproachThe graph approach is a strong tool but it's not perfect. If a graph is not provided and it seems sufficiently annoying or difficult to make your own graph, don't bother. Additionally, there can be cases where the graph does not seem to indicate an issue even though one exists. For example, examine the graph of $f(x) = x^{3/5}$: At $x=0$, there is a vertical tangent line, and vertical tangent lines are places that are smooth-looking yet not differentiable. This is one reason that, while graphs are extremely helpful for identifying differentiability issues, we must always be able to rely on the mathematical limit definition.
Remember! Continuity is a prerequisite for differentiability. Anywhere that a function is not continuous is also a place where that function is not differentiable. However, the converse is not always true - it is possible for a function to be continuous at a place where it is not differentiable. As an example, consider the power function above with the instantaneous vertical tangent, or the function $f(x) = |x|$.
The Proper ApproachThe most important thing you can take away from this lesson is knowing how to algebraically find un-differentiable function values for a given function. Unlike the graph approach which is based on visual cues and reasonable suspicion, the "proper" approach is entirely based in Algebra and Calculus steps.Let's say you are given a function $f(x)$, and you are asked to determine where it is differentiable. Since differentiability is a synonym for having $f'(x)$ be continuous, start by finding $f'(x)$, and then see if that result has any discontinuities. Nearly 100% of the time that means looking for domain problems, like division by zero or square roots of negative numbers. If it is a piecewise function, then the interval endpoints are also possible sources of trouble.

Example 1Find any values of $x$ that cause$$f(x) = \sqrt{|x|}$$to be not differentiable.$\blacktriangleright$ As mentioned previously, a graph is helpful in identifying possible places of concern. Right away, it is clear that $x=0$ is a concern. Let's make sure that our suspicion is correct by applying the limit definition of differentiability. We'll start with the left side.$$\lim_{x \to 0^{-}} \frac{d}{dx} \, \sqrt{|x|}$$Since $x$ is negative in this case, $\sqrt{|x|}$ is really $\sqrt{-x}$ so that the result is positive.$$\lim_{x \to 0^{-}} \frac{d}{dx} \, \sqrt{-x} = \lim_{x \to 0^{-}} \frac{-1}{2\sqrt{-x}}$$$$= \frac{-1}{2\sqrt{-0^{-}}}$$$$\longrightarrow -\infty$$Now let's examine the right ride.$$\lim_{x \to 0^{+}} \frac{d}{dx} \, \sqrt{|x|}$$From the right side, $|x|$ is already a positive number, and can be replaced with $x$:$$\lim_{x \to 0^{-}} \frac{d}{dx} \, \sqrt{x} = \lim_{x \to 0^{+}} \frac{1}{2\sqrt{x}}$$$$= \frac{1}{2\sqrt{0^{+}}}$$$$\longrightarrow +\infty$$Without a real number tangent slope, the derivative cannot exist at $x=0$, and therefore, this function is not differentiable at $x=0$.

Let's look at one more example, and more plainly see that the location with a differentiability issue can be quickly identified using $f'(x)$, and without requiring a graph.

Example 2Determine the location of non-differentiable points on the function $f(x) = x^{3/5}$.$\blacktriangleright$ This is a function we discussed with a pure graph approach, but let's now try the algebraic approach.$$f(x) = x^{3/5}$$$$\longrightarrow f'(x) = \frac{3}{5x^{2/5}}$$We can see that the function $f'(x)$ will have issues at any $x$ value that makes the denominator zero, namely, $x=0$ in this case. Since $f'(x)$ is undefined at $x=0$, we can say that the function is not differentiable at $x=0$.The key takeaway from the proper approach is that we have a rigorous and quick way to look for differentiability problems. If you're given a function and need to know the points where it is not differentiable, take its derivative and look for domain issues.
Pro Tip The meat and potatoes of what we need to know about differentiability is two quick ideas:
• Differentiability refers to smoothness, which happens when $f'(x)$ is continuous.
• To find differentiability issues for a given function, find $f'(x)$ and look for domain issues.

## Typical Functions

As with continuity and other core derivative skills, the core function families have common and well-behaved properties that we should know and understand.
Common Function DifferentiabilityThe following functions are continuous and differentiable everywhere on their domain:
• Polynomial
• Trigonometric
• Exponential
• Logarithmic
• Rational
Warning! This list is not quite as complete as the list for functions that are continuous. Radical functions are almost differentiable everywhere, but usually have a vertical tangent at their "end".Additionally, power functions of the form $f(x) = x^p$ where $p$ may be non-positive or non-integer may contain vertical tangents or cusps. So even though their derivatives are computed the same as polynomial terms (using the power rule »), do not assume they are everywhere differentiable.Don't forget that when we use the word polynomial, it specifically means that the exponent on any term is a positive integer only - no negative or fraction exponents.

## Mr. Math Makes It Mean

There are a few ways teachers can ask about differentiability in less-than-direct ways.Combinations of FunctionsFor any type of function combination, whether it's arithmetic of functions or composition of functions, the best plan is to consider the final product and not focus on the individual functions that are being used. Sometimes it will be enough to inspect the combination function, but we always have the option of explicitly taking its derivative and looking for continuity issues.The only thing about the foundational functions that we need to pay attention to is the domain restrictions, because we can't eliminate domain problems via composition. This next example will remind us of that.

Example 2For $f(x) = \sqrt{x^2 -2x -15}$ and $g(x) = x^2-4$, determine whether $f(g(x))$ and $g(f(x))$ are differentiable everywhere, and if not, state the intervals on which they are.$\blacktriangleright$ The best practice is to just find the functions and look for issues. Don't start looking at each function separately.For $f(g(x))$:$$f\left(g(x)\right) = \sqrt{\left(x^2-4\right)^2 -2\left(x^2-4\right) - 15}$$$$= \sqrt{x^4-10x^2 +9}$$When it comes to radical functions, we need to remember that they are not defined on intervals that make the expression under the root negative. Additionally, the function will not be differentiable at the value that makes the expression under the root zero - the function is defined there but the tangent slope is vertical.$$=\sqrt{\left(x^2-9\right)\left(x^2-1\right)}$$$$=\sqrt{(x+3)(x-3)(x+1)(x-1)}$$Using sign analysis (which we sometimes call a sign chart or sign diagram), we can tell that $x^4 -10x^2 +9$ is negative on the intervals $(-3,-1)$ and $(1,3)$. On those intervals, the function is undefined, so of course it is not differentiable on those intervals either. The function is not differentiable right at the points where $x=-3$, $-1$, $1$, or $3$. Therefore our answer is:$f(g(x))$ is differentiable on $(-\infty,-3) \cup (-1,1) \cup (3,\infty)$.Note that we didn't really need to take the derivative of $f(g(x))$ to find these issues. We could have, but just using familiarity with radical functions is a nicer approach because$$\frac{d}{dx} \, \sqrt{x^4 -10x^2 +9}$$$$=\frac{2x^3 -10x}{\sqrt{x^4 -10x^2 +9}}$$which isn't fun to look at.For $g(f(x))$:$$g\left(f(x)\right) = \left(\sqrt{x^2-2x-15}\right)^2 -4$$$$=x^2-2x-19$$This result is a polynomial function, and we know that polynomial functions are differentiable everywhere. However, like continuity, we aren't allowed to include any $x$ values in the domain that we couldn't include in the original functions. Therefore, since $g(x)$ isn't defined for $x \in (-3,5)$, $g(f(x))$ isn't either and is therefore also not differentiable. It is however differentiable at $-3$ and $5$ because the derivative of $g(f(x))$ is defined at those points.

## Unknown Coefficients Problems

Teachers love to test your knowledge of differentiability with piecewise functions. One common way they do this is to ask you to find the unknown coefficients in a piecewise function such that the function is differentiable at the endpoints of the piecewise interval.

Example 3Find $a$ and $b$ such that the following function $f(x)$ is continuous and differentiable. $\blacktriangleright$ From Algebra, we should know that having two unknowns in our problem implies that we need to solve two simultaneous equations. First, we need this function to be continuous. In order for that to be true, the left-side function near $x=2$ must equal the right-side function near $x=2$.$$ax^2 - bx + 1 = ax - b - 1$$but since we need this true at $x=2$, replace $x$ with $2$:$$4a - 2b + 1 = 2a - b - 1$$$$\Rightarrow 2a - b = -2$$Now, to get the other equation, use differentiability. The problem requires that the function is differentiable at $x=2$, which means the tangent slope on each side near $x=2$ must be the same. To solve for this, we set the derivatives of each side's function equal to one another, and then let $x$ equal $2$.$$\frac{d}{dx} \, ax^2 - bx + 1 = \frac{d}{dx} \, ax - b - 1$$$$\Rightarrow 2ax - b = a$$$$\Rightarrow 2a(2) - b = a$$$$\Rightarrow 4a - b = a$$$$\Rightarrow 3a - b = 0$$Now, with two equations, we can solve a system and figure out what each $a$ and $b$ need to be so that the problem's requirements are true.$$2a - b = -2$$$$3a - b = 0$$Subtracting the second equation from the first:$$-a = -2$$$$\longrightarrow a=2$$Now, if $a=2$, then via either equation in the system, we know that $b=6$.

## Put It To The Test

Make sure you can do the various types of problems that teachers like to ask for on quizzes and tests!

Example 4Use the given graph of $f(x)$ and determine where $f$ is not differentiable. Show solution
$\blacktriangleright$ Since the graph has a discontinuity at $x=8$, we know right away that $f$ is not differentiable there either. Furthermore, using visual inspection, it seems like $f(x)$ is not differentiable at $x=4$. Finally, it looks like $f$ is differentiable at $x=2$, because even though the graph definition seems to change piecewise definitions, the slope looks like it has a horizontal tangent on each side of the join point.

In Examples 5 - 9, determine all $x$ values for which the function is not differentiable.

Example 5$$\frac{x+3}{x-1}$$
Show solution
$\blacktriangleright$ Rational functions are differentiable everywhere on their domain. Since this function only has a domain issue at $x=1$, we can be certain that this is the only place that the function is not differentiable.A more rigorous proof of this fact comes from the fact that$$\frac{d}{dx} \, \frac{x+3}{x-1}$$$$=-\frac{4}{(x-1)^2}$$which is defined everywhere except for $x=1$. The translation of this is that the derivative, which is the tangent slope function, has a value at every value of $x$ except $1$ and therefore we know the only place a derivative doesn't exist is $x=1$.

Example 6$$\sin\left(x^2\right)$$
Show solution
$\blacktriangleright$ This function belongs to one of the function families that we know is automatically continuous and differentiable everywhere on its domain. Furthermore, $\sin\left(x^2\right)$ has no domain issues, so we can be sure that it is indeed continuous everywhere.If you need to show this more rigorously based on your instructions for the problem, simply take the derivative and see that it has no domain issues either.$$\longrightarrow \frac{d}{dx} \, \sin\left(x^2\right)$$$$= 2x \sin\left(x^2\right)$$

Example 7$$|x-2|$$
Show solution
$\blacktriangleright$ By now you're hopefully passingly familiar with the absolute value function, and enough to know that it will not be differentiable at the "corner" of the v shape since it is not smooth (or more rigorously, each side of the corner has a different slope).Based on its graph and needing only to answer the question, we know this is going to occur at $x=2$, and that the function $|x-2|$ is not differentiable at that point. If you need to show this with more rigor, re-write this absolute value function as a piecewise function. We can see that the derivative of each definition is $1$ and $-1$ respectively, and that the function has one slope on one side of $x=2$ and the other slope on the other side of $x=2$. Therefore, the limit of $f'(x)$ does not exist because each one-sided limit does not agree with the other, thus showing with rigor that the function $|x-2|$ is not differentiable.

Example 8$$\sqrt{x^2-5x-6}$$
Show solution
$\blacktriangleright$ Let's figure this one out without a graph. Start with the derivative of the function using the Chain Rule:$$\frac{d}{dx} \, \sqrt{x^2-5x-6}$$$$\longrightarrow \frac{2 x - 5}{2 \sqrt{x^2 - 5 x - 6}}$$Now the question is, where does the derivative have domain issues? We can't let the denominator equal zero, nor can we let it be negative. The best way to examine the expression under the radical is with sign analysis.$$x^2 - 5x - 6 = (x-6)(x+1)$$When is $(x-6)(x+1)$ negative? Look to the three intervals that the two zeros break the number line into. This expression is zero when $x=6$ or $x=-1$.With some test points, we can quickly verify that left of $x=-1$ and right of $x=6$, the expression is positive, and in between $-1$ and $6$, the expression is negative. We also cannot allow $x$ to be anything that makes the denominator zero, so $x=-1$ and $x=6$ are not allowed either. Putting this altogether, the domain of the derivative is$$x \in { \mathbb{R}, \,\,\, | x < -1, \,\, x>6}$$which will also be the places where the function is differentiable.

Example 9$$\frac{2x^2 + 7x - 4}{x^2 - 2x - 24}$$
Show solution
$\blacktriangleright$ After some fast analysis on this rational function, we'll see that there is a removable discontinuity at $x=-4$. Therefore, if we want to understand the behavior of this function at any other point beside when $x=-4$, we can cancel those matching factors and examine what we're left with.$$\frac{(2x-1)(x+4)}{(x+4)(x-6)}$$$$\longrightarrow \frac{2x-1}{x-6}$$It is important to remember that we can pretend the factors canceled to see what simpler function we are left with, which will be easier to understand and graph, but the original function will still have those factors present. We cannot ignore the fact that the original function is undefined at that point, even though the "problem" factors cancel out.Examining the reduced function, we can see that the only remaining domain issue we have is $x=6$, at which point the function is not defined. Since, in accordance with the definition in this lesson, rational functions are differentiable everywhere on their domain, we know we'll have differentiability guaranteed for this function for all real number inputs except $x=-4$ and $x=6$. Both $x$ input values are domain errors, even though one of them cancels out.
Warning! Don't include a domain problem that "cancels out" in your solution. If it breaks the function as originally written, then it's not valid.

Example 10When graphed, power functions often appear to be smooth everywhere. Determine whether the function $f(x) = x^{1/3}$ is differentiable everywhere on its domain by considering both a graph and a limit approach.
Show solution
$\blacktriangleright$ With a graph, it may not immediately be clear if we are looking at any problems or issues. The limit approach, however, will give us definitive results.$$f(x) = x^{1/3}$$$$f'(x) = \frac{1}{3x^{2/3}}$$Using the limit approach to differentiability, we are essentially looking for places where the derivative function $f'(x)$ has domain issues. Clearly, $x$ cannot be $0$ in the derivative, else the derivative function will be undefined. This also makes sense graphically - at $x=0$, the graph exhibits the behavior of a vertical tangent line, which is a graphical hint that differentiability will fail even when continuity doesn't.This is the only domain issue of the derivative, so it is the only place of concern. We can safely say that $f(x) = x^{1/3}$ will be differentiable everywhere except $x=0$.

Example 11Show that the following function is continuous but not differentiable at $x=1$. Show solution
$\blacktriangleright$ To show that this function is continuous at $x=1$, we must show that the limit of the function from each side of $x=1$ is equal to the actual function value $f(1)$.$$f(1) = 3$$$$\lim_{x \to 1^{-}} \, f(x) = (1)^2 + 2 = 3$$$$\lim_{x \to 1^{+}} \, f(x) = (1) + 2 = 3$$This criteria checks out.In order to be differentiable, we need the derivative on each side of $x=1$ to approach the same tangent slope.$$\lim_{x \to 1^{-}} \, f'(x) = 2x \rvert^{x=1} = 2$$$$\lim_{x \to 1^{+}} \, f'(x) = 1 \rvert^{x=1} = 1$$Since these answers do not agree, the function is not differentiable at $x=1$, even though it is continuous.

Lesson Takeaways
• Understand differentiability as both an intuitive measure of smoothness and as a limit definition
• Know how to use graphs to evaluate differentiability and know the limitations of using a graph based approach
• Be able to evaluate where a function is and is not differentiable using the Calculus approach without graphs.
• Know the differentiation properties of common function families
• Approach combinations of functions properly, examining the final result which remembering domain restrictions on either standalone function
• Be ready to answer questions about piecewise functions, including missing coefficient problems
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