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### Course: Calculus, all content (2017 edition)>Unit 2

Lesson 8: Differentiability

# Differentiability at a point (old)

An older video where Sal finds the points on the graph of a function where the function isn't differentiable. Created by Sal Khan.

## Want to join the conversation?

• this video feels out of place...what does differentiable mean?
• A function is differentiable at a point when there's a defined derivative at that point. This means that the slope of the tangent line of the points from the left is approaching the same value as the slope of the tangent of the points from the right.

Still confused?
In all the examples he gave, the points where the functions weren't differentiable were where the function changed course. This means the speed it was going at completely changed. From the left side of the point, the function was going at one speed, but from the right side of the point, the function was going at a different speed. Since the function had different speeds from both sides, there was no defined speed at the point where the change happened. The function isn't differentiable at the point where the change happens.

I hope this explains what differentiable means well!
• can this be proven mathematically? Like Can you write an equations showing how certain point will not be able to find a derivative?
• yeah, using formal definition for derivative we will prove that the 2 limits are different if we approach from neg or pos direction
• So, essentially, a function is not differentiable at points of discontinuity?
• Right. A differentiable function is always continuous, but the inverse is not necessarily true. A derivative is a shared value of 2 limits (in the definition: the limit for h>0 and h<0), and this is a point about limits that you may already know that answers your question. At points of discontinuity of f(x) the derivative, which is a shared value of 2 limits (the derivative from the right and the derivative from the left) can only be calculated from one side (at best). So there can't be a shared value. Exactly like what happens with a (bounded from above) continuous function on, for example, [0,1[ : The left limit in x=1 will exist, but the limit won't, because you can't "approach it" from the right, and by consequence you can't check that value.

You should also check this out: http://en.wikipedia.org/wiki/Weierstrass_function. This function is continuous everywhere, but differentiable nowhere.
• Dumb question , but It's really confusing me. How did sal find that the slope was (3.5) at .
• My friend and I were discussing about an interesting yet confusing question.
The question is that If we divide the function in the graph into 3parts: y1(x<= -2), y2( -2<x<=3), and y3(x>3), are the points (-2,-3) and (3, 4.5 approximately) still not differentiable?
I voted for still not but my friend voted for that they are differentiable in this occasion.
What would be the legitimate answer?
• Nope, not differentiable. The points do have "one-sided" derivatives but because they're not the same on both sides, neither point has a derivative.
• I can get the idea about differentiability from this video. However, it seems there isn't any video to explain this problem clearer. How can we apply it in real situation?
• Classic example: imagine someone is running in a straight line, with variable speed. We construct a function: the x-axis represents time, and the y-value represents his displacement (the distance away from the starting point). If you want to know his exact speed at a given moment (dx/dt, the ratio of the change in displacement and the change in time as the change in time approaches zero), you calculate the (first) derivative of this function at a given time t.
• SUPPOSE, somehow we got a tangent line that is vertical, i.e. of type x=k, where k is some constant. Then will the value of derivative be defined there? Also, in REALITY, is there any such possible case?
• Yes, for example a graph of circle would give a vertical tangent line at the maximum and minimum x-values.
• So when you say when a function is not differentiable, is it the same thing as saying where the graph is not continuous?
• Not necessarily. Consider f(x) = |x|. f(x) is continuous on (-infinity, infinity), but it's not differentiable at x = 0.
• So is differentiability essentially continuity of the derivative of a function?
• The derivative of a function need not be continuous. For instance, the function `ƒ: R → R` defined by `ƒ(x) = x²sin(1/x)` when `x ≠ 0` and `ƒ(0) = 0`, is differentiable on all of `R`. In particular, `ƒ` is differentiable at `0` (in fact, `ƒ'(0) = 0`), but the derivative `ƒ'` of `ƒ` is not continuous at `0`.

However, if we consider functions of a complex variable, this is indeed the case. More precisely, if a function `ƒ` is complex differentiable on an open subset `Ω` of the complex plane, we say that `ƒ` is holomorphic on `Ω`. If `ƒ` is such a function, then `ƒ` has derivatives of all orders on `Ω`! In particular, the derivative `ƒ'` of `ƒ` is continuous on `Ω`. This is in contrast to the real case, and we see that the notion of being holomorphic is stronger than the notion of being real differentiable.
• so in short a point isnt differntiable if a slope on the left and right of that isn't the same??
• Yes, you are correct! The definition of derivative
lim h->0 of [ f(x+h) - f(x) ] /h
Implies that you should be able to take the limit from both sides, and they should both be equal. This is equivalent to saying the slopes on both sides are equal.

## Video transcript

Consider f which is defined for all real numbers. At what arguments x is f of x not differentiable? So to think about that, I'm actually going to try to visualize what f prime of x must look like. So I'm going to do f prime of x in this purple color. So if we look at f of x right over here, it looks like its slope is pretty much consistently negative 2 over this interval between x equals, I guess, it's like negative 8 and 1/2 all the way up to x equals negative 2. It looks like the slope is a constant negative 2. So if I were to draw its derivative, its derivative would look something like this. Its derivative looks something like this. But then something interesting happens at x equals negative 2. Right as we cross x equals negative 2, it looks like the slope goes from being negative to being positive. And it looks like right out the get go, if I were to estimate the slope of its tangent line, it starts changing. It's not a line anymore. It's a curve. The slope of the tangent line right at this point looks like it's around-- I don't know-- it looks like it's around 3 and 1/2. Because if I were to draw a tangent line right over here, it looks like if I move 1 in the x direction, I move up about 3 and 1/2 in the y direction. So I'm just trying to, obviously, estimate it. So it looks like the slope goes up to 3 and 1/2 right when I cross that point. And then the slope becomes lower and lower and lower all the way until I get to this point right over here, all the way until I get to x equals 2. And it looks like it continues to get lower all the way until you get to x equals 3. So it looks like the slope of the line is-- it looks like it's getting lower at a constant rate, I guess I could say. So it looks like it's doing something like this over this interval. But then right as x crosses 3, this becomes a flat line. The slope is 0 here. So right as x crosses 3, the slope becomes 0. So we immediately see there are points where it looks like the slope jumps. And at these points we really don't have a defined derivative. The slope jumps there as well. And so at what arguments is f not differentiable? Well, it's not differentiable when x is equal to negative 2. When x is equal to negative 2, we really don't have a slope there. Remember, when we're trying to find the slope of the tangent line, we take the limit of the slope of the secant line between that point and some other point on the curve. If we did that as we approached from the left, it looks like the derivative is negative 2. If we do that from the right, it looks like the derivative is something like positive 3 and 1/2. And so we're not getting the same limit of the secant line as we approach from the left and as we approach from the right. And the same thing is happening at x is equal to 3. At x equals 3, as we approach from the left, the slope looks like it is decreasing. It is approaching-- I don't know-- maybe around negative 1. But as we approach from the right it looks like the slope is 0. So we do not have the same limit of the secant slope as we approach from the left- and right-hand sides. So at both of these points we see the derivative jump, and it looks like f of x is not differentiable.