A derivative exists at a point only if the limit defining it exists. Some perfectly ordinary graphs fail that test.
Step 1: Let's Learn
Read it, or press Listen and follow the words.
Corners
At the corner of y = |x| the slope from the left is −1 and from the right is 1. They disagree, so no single tangent exists.
Cusps
A cusp is a sharper version of the same failure, where the two sides approach steepness in opposite directions.
Vertical tangents
At a vertical tangent the slope would be undefined, since a vertical line has no slope.
Differentiable implies continuous
Every differentiable function is continuous. A break in the graph makes the defining limit fail immediately.
The reverse is false
y = |x| is continuous everywhere yet not differentiable at zero. Continuity is necessary but nowhere near sufficient.
Where a function has no slope
A function is differentiable at a point if the difference quotient limit exists there. Corners, cusps, vertical tangents and discontinuities all destroy it, each for a visible reason.
A corner has two different slopes
At the corner of y = |x| the slope from the left is −1 and from the right is +1. The one-sided limits disagree, so no derivative exists. The function is continuous there and still not differentiable.
Differentiable implies continuous
A function with a derivative at a point must be continuous there, but the converse fails — |x| is the standard counterexample. Differentiability is strictly the stronger condition.
Smoothness is what differentiability means
A differentiable function locally looks like its tangent line: zoom in far enough and the curve becomes straight. That local linearity is the real content of differentiability and the basis of linear approximation.
Step 2: Try It Yourself
Tap and try it out.
Step 3: Watch an Example
One step at a time.
Watch Diego Test y = |x| at Zero
Diego checks whether the absolute value function has a derivative at x = 0.
- Step 1
For x below zero the function is −x, so its slope there is −1.
Step 4: Your Turn
Practice makes it stick.
The Corner
Problem 1 of 2
For y = |x|, what is the slope just to the left of zero?
The Implication
Problem 2 of 2
A function is differentiable at x = 3. Must it be continuous there? 1 for yes, 0 for no.
Where the Slope Fails
1 of 8
For y = |x|, the slope just right of zero?
2 of 8
A function is continuous at x = 2. Must it be differentiable there? 1 or 0.
3 of 8
y = |x − 4|. At which x is it not differentiable?
4 of 8
A graph jumps at x = 1. Is it differentiable there? 1 or 0.
5 of 8
y = x². Is it differentiable at x = 0? 1 or 0.
6 of 8
At a vertical tangent, is the derivative defined? 1 or 0.
7 of 8
Which situations block differentiability?
8 of 8
y = |x + 7|. At which x is it not differentiable?
Step 5: Quick Check
Show what you know.
Question 1 of 2
y = |x − 2|. At which x is the derivative undefined?
Question 2 of 2
Which statement is true?
What You Learned
- A derivative fails to exist at corners, cusps, vertical tangents and breaks.
- Every differentiable function is continuous.
- Continuity does not guarantee differentiability.