Connecting Differentiability and Continuity: Determining When Derivatives Do and Do Not Exist
Why this matters
Imagine you're on a cross-country road trip from Boston to Seattle. For most of the trip, the highway is smooth and easy to drive. We could describe the "slope" of the road at any given point. But what happens if you hit a snag?
Maybe you come to a river where the bridge is out. The road just stops, and there's a huge gap. You can't drive across it. The road isn't continuous.
Or, maybe the road is fully connected, but you get to a part in downtown Chicago with an impossibly sharp, 90-degree turn. You can't drive through it without stopping and turning the wheel completely. The road is connected, but it's not smooth.
In calculus, we call a "connected" road continuity and a "smooth" road differentiability. Today, we're going to explore their exact relationship. You'll learn why you can't have a smooth road with a gap in it, but you can have a connected road that’s too sharp to be called smooth.
Concept overview
flowchart TD
A[Start with a function f(x) at point x=c] --> B{Is f(x) continuous at x=c?};
B -- No --> C[Not Differentiable];
B -- Yes --> D{Is the graph smooth at x=c?};
D -- No (Corner, Cusp, or Vertical Tangent) --> C;
D -- Yes --> E[Differentiable];
Core explanation
Hello there! I'm Saavi, and I'm excited to walk you through one of the most fundamental ideas in calculus. Let's clear up the connection between being differentiable and being continuous.
At its heart, the derivative tells you the instantaneous rate of change, or the slope of the tangent line at a single point. For a derivative to exist at a point x = c, we must be able to zoom in on that point and have the function look like a straight line.
Differentiability Implies Continuity
Let's start with the golden rule of this topic: If a function is differentiable at a point, it must be continuous at that point.
Think back to our road trip. If a road is "differentiable" at a certain spot—meaning it's smooth enough to have a well-defined slope—it must, by definition, exist at that spot. You can't talk about the smoothness of a piece of highway if there's a giant chasm where the highway should be.
In mathematical terms, for a derivative f'(c) to exist, the function f(c) has to be there. No holes, no jumps, no vertical asymptotes at that point. The function must be continuous at c.
This also means that if a point isn't even in the domain of a function, you can't even begin to ask if it's differentiable there. If f(5) is undefined, f'(5) is also certainly undefined.
So, the big takeaway is: Differentiability is a stronger condition than continuity. It demands more from the function. It's like saying that to be an eligible voter in the US, you must be a citizen. But not all citizens are eligible voters (some are too young). Differentiability is the "eligible voter," and continuity is the "citizen." All differentiable functions are continuous, but not all continuous functions are differentiable.
Continuity Does NOT Imply Differentiability
Our road can be fully connected, but still have features that make it impossible to define a single, unique slope at a point. There are three classic ways this happens:
1. A Corner or Cusp
Imagine you're walking along the graph of f(x) = |x|. The graph looks like a perfect "V" with its point at the origin. The function is clearly continuous at x = 0—you can draw the whole graph without lifting your pencil.
But what's the slope at x = 0?
- If you approach from the left, the slope is a constant -1.
- If you approach from the right, the slope is a constant +1.
Since the slope from the left (-1) does not equal the slope from the right (+1), we can't agree on a single tangent line at the origin. Therefore, the derivative does not exist at x = 0. This sharp point is called a corner. A cusp is similar but more pinched, like the point on a bird's beak.
f(x) = |x|
^
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| /
| /
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<-----+----->
|At x=0, the function is continuous but not differentiable.
2. A Vertical Tangent Line
Consider the function f(x) = x^(1/3) (the cube root of x). This graph is continuous everywhere. But look at what happens at x = 0. The graph goes almost vertical for a moment.
If you try to draw a tangent line at x = 0, it would be a perfectly vertical line. What's the slope of a vertical line? It's undefined! Since the slope is undefined, the derivative does not exist at x = 0.
This is like our road trip hitting a section that goes straight up a cliff face. The road is "connected," but its slope is infinite, which we treat as undefined.
3. Oscillation (Less Common, but Good to Know)
This is a wild one. Some functions, like f(x) = x * sin(1/x) near x=0, wiggle so rapidly as they approach a point that you can't pin down a single slope. The function is continuous, but the slope bounces around infinitely fast, so a derivative can't be defined. You likely won't see this on an exam, but it's another way a continuous function can fail to be differentiable.
Summary of the Relationship
- Is the function continuous at the point?
- No?Then it's not differentiable. Stop here.
- Yes?Proceed to the next question.
- No?
- Is the function smooth at the point (no corner, no cusp, no vertical tangent)?
- No?Then it's not differentiable.
- Yes?Then it is differentiable.
- No?
Remembering this two-step check will save you a lot of points. Always check for continuity first!
Worked examples
Let's put this theory into practice with a few examples you're likely to see.
The Piecewise Function Corner
Problem: Consider the function f(x) defined as:
f(x) = { x^2 + 1, if x <= 1
{ 2x, if x > 1
Is the function f(x) differentiable at x = 1?
Solution:
This is a two-part test. First, we must check for continuity. Then, if it's continuous, we check if the slopes match.
Step 1: Check for Continuity at x = 1 For a function to be continuous at a point, the limit from the left, the limit from the right, and the function's value must all be equal.
- Function value
f(1)uses the top rule (x <= 1), sof(1) = (1)^2 + 1 = 2. - Limit from the left
lim (x->1⁻) f(x)also uses the top rule.lim (x->1⁻) (x^2 + 1) = 2. - Limit from the right
lim (x->1⁺) f(x)uses the bottom rule.lim (x->1⁺) (2x) = 2(1) = 2.
Since the left limit, right limit, and function value are all 2, the function is continuous at x = 1. We can proceed to the next step.
Step 2: Check for Differentiability at x = 1 Now, we need to see if the slopes from the left and right match up. We do this by finding the derivative of each piece.
- Derivative of each piece
- For
x < 1, the derivative isd/dx (x^2 + 1) = 2x. - For
x > 1, the derivative isd/dx (2x) = 2.
- For
- Slope from the left (at x = 1)We plug
x=1into the first derivative:2(1) = 2. - Slope from the right (at x = 1)The derivative is a constant
2.
The slope from the left is 2, and the slope from the right is 2. Since they are equal, the function is differentiable at x = 1.
Wait, what? This is a great example of why you must do the math. It might look like a corner, but the parabola x^2+1 and the line 2x meet so perfectly at x=1 that their slopes match exactly at that point, creating a smooth transition.
Let's tweak it slightly to see a failure. If the second piece was 3x - 1, it would still be continuous at x=1 (since 3(1)-1 = 2), but the derivative from the right would be 3. Since 2 != 3, that function would have a corner and not be differentiable at x=1.
The Absolute Value Corner
Problem: Show that g(x) = |x - 4| is not differentiable at x = 4.
Solution:
Step 1: Check for Continuity at x = 4
The absolute value function is continuous everywhere. You can plug in x=4 to get g(4) = |4 - 4| = 0. The limits from the left and right also approach 0. So, it's continuous.
Step 2: Check for Differentiability at x = 4
Let's rewrite g(x) as a piecewise function, which is the key to working with absolute values.
g(x) = { -(x - 4), if x < 4
{ (x - 4), if x >= 4
This simplifies to:
g(x) = { 4 - x, if x < 4
{ x - 4, if x >= 4
Now, find the derivative of each piece:
- Slope from the left (x < 4)
d/dx (4 - x) = -1. - Slope from the right (x > 4)
d/dx (x - 4) = 1.
The slope approaching from the left is -1, and the slope approaching from the right is 1.
Try it yourself
Ready to try a couple on your own? Take your time and apply the two-step process.
Problem 1:
Find the values of a and b that make the function h(x) differentiable for all real numbers.
h(x) = { ax^2, if x <= 2
{ 2x + b, if x > 2
- Set the two function pieces equal to each other at
x = 2to ensure continuity. - Set the derivatives of the two pieces equal to each other at
x = 2to ensure smoothness.
Problem 2:
The graph of a function k(x) is shown below. It has a sharp corner at x = -3 and a removable discontinuity (a hole) at x = 5. List all x-values where k(x) is not differentiable and explain why for each.
(Imagine a graph with a "V" shape at x=-3 and a smooth curve with an open circle at x=5).
In simple terms, this lesson explains the relationship between a function being "smooth" (differentiable) and "unbroken" (continuous). A function must be unbroken to be smooth, but just being unbroken isn't enough to guarantee smoothness.
f(x) = |x|
^
| /
| /
| /
| /
<-----+----->
|- FUN-2.A: Explain the relationship between differentiability and continuity.
- FUN-2.A.1
- If a function is differentiable at a point, then it is continuous at that point. In particular, if a point is not in the domain of f, then it is not in the domain of f'.
- FUN-2.A.2
- A continuous function may fail to be differentiable at a point in its domain.
flowchart TD
A[Start with a function f(x) at point x=c] --> B{Is f(x) continuous at x=c?};
B -- No --> C[Not Differentiable];
B -- Yes --> D{Is the graph smooth at x=c?};
D -- No (Corner, Cusp, or Vertical Tangent) --> C;
D -- Yes --> E[Differentiable];
Read what Saavi narrates
Hi there, it's Saavi from Shrutam. Let's talk about one of those ideas in calculus that seems tricky at first, but is actually pretty logical once you see it.
Imagine you're on a road trip. For the road to be considered "smooth," it first has to be "connected," right? You can't have a smooth road if there's a giant hole in it. But just because the road is connected, doesn't mean it's smooth. You could hit a really sharp, pointy turn.
That's the whole idea behind differentiability and continuity. A "smooth" function is differentiable. A "connected" function is continuous.
So, the big rule is this: for a function to be smooth, or differentiable, it must first be connected, or continuous. But just because it's continuous doesn't mean it's differentiable.
Let's look at a classic example: the function g of x equals the absolute value of x minus 4.
The graph of this function is a "V" shape, with the point of the V at x equals 4.
It's definitely continuous... you can draw the whole V without lifting your pencil. But is it differentiable at x equals 4?
To find out, we have to look at the slope from the left and the slope from the right.
If you approach the point from the left side, the line has a slope of negative one.
But if you approach from the right side, the line has a slope of positive one.
Since negative one does not equal positive one, we can't define a single slope *at* that sharp point. So, the function is not differentiable there.
This brings up a really common mistake: assuming that if a function is continuous, it must be differentiable. Please don't fall into that trap! Always remember that V-shape. It's connected, but it's not smooth at the point. Continuity is necessary, but it's not enough.
You're building a really strong foundation here. Keep asking these "why" questions, and you'll find that calculus starts to make a lot of sense. You've got this.
This is the main point of the lesson. Continuity is necessary, but not sufficient. Functions can have sharp corners or vertical tangents and still be continuous.
Always remember the one-way relationship: Differentiability -> Continuity. Think of the `f(x) = |x|` example; it's continuous at `x=0` but not differentiable there.
You can waste a lot of time finding derivatives of a piecewise function, only to find out it wasn't even continuous to begin with. If it's not continuous, it can't be differentiable.
Always perform the 3-part continuity check at the point in question *first*. If it fails, you're done. The function is not differentiable.
A derivative of zero means the tangent line is horizontal. This is a perfectly valid, existing derivative. A non-existent derivative means you can't define a unique tangent line at all (e.g., at a corner).
At the top of a smooth hill on a graph (like `y = -x^2` at `x=0`), the derivative is 0. At the point of a "V" shape (like `y = |x|` at `x=0`), the derivative Does Not Exist (DNE).
This only proves continuity. To prove differentiability, you must also show that the derivatives (the slopes) are equal at the boundary.
For a piecewise function at a boundary point, perform two checks: 1) Are the function pieces equal? (Continuity). 2) Are the derivatives of the pieces equal? (Smoothness). You need a "yes" on both.