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## Calculus 1

### Course: Calculus 1 > Unit 2

Lesson 5: Differentiability- Differentiability and continuity
- Differentiability at a point: graphical
- Differentiability at a point: graphical
- Differentiability at a point: algebraic (function is differentiable)
- Differentiability at a point: algebraic (function isn't differentiable)
- Differentiability at a point: algebraic
- Proof: Differentiability implies continuity

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# Differentiability and continuity

Defining differentiability and getting an intuition for the relationship between differentiability and continuity.

## Want to join the conversation?

- People say that x^2 is differentiable at x=0. Why is this? It would seem like it is not differentiable because it seems like the absolute value function.(24 votes)
- x^2 is a parabola centered at the origin....If you take its derivative you get 2x, therefore the derivative of f(x) at 0 would be equal to 0... or you can write as f'(0) = 0....It is a parabola you do not have a hard corner where you would end up with an infinite number of slopes crossing that point......(37 votes)

- At5:52, why is the expression approaching negative infinity?(17 votes)
- Think of slopes as rise over run. In this case it is negative infinity because you start of with a change in x which is negative (c being bigger than x) and a change in y which is positive (being f (x) larger than f (c))..... therefore dy/dx is negative, and as x approaches c the limit of the function from the left is equal to negative infinity.....(19 votes)

- So differentiability means that a certain point on a function has only 1 possible tangent line with a specific slope? What about the tops or bottoms of curves? It looks like they could have varying tangent lines, so would they still be differentiable?(5 votes)
- Tops and bottoms of curves have a slope of 0, imagine driving a car and looking perfectly parallel to the ground. Your vision is the tangent line. As you go up a hill the tangent is constantly changing, but there's still only "one" true tangent line at any exact point. True, if you move just .000001 inches then the tangent may change but that's not really the point.

Yes, from either side of the hill/curve the tangents have different slopes, but as they each approach that "top" point, they should become equal to one another, again where the slope is equal to 0.

There are some exceptions, especially for a function that has a very sharp curve, like y = |x|, these slopes one either side are completely opposite (-1 and 1), and so at the "bottom" there is no tangent. The reason is because for a function the be differentiable at a certain point, then the left and right hand limits approaching that MUST be equal (to make the limit exist). For the absolute value function it's defined as:

y = x when x >= 0

y = -x when x < 0

So obviously the left hand limit is -1 (as x -> 0), the right hand limit is 1 (as x -> 0), therefore the limit at 0 does not exist!

For other functions that have more gentle curves then you get a more gradual shift toward the same limit near the top/bottom of a curve, mainly they approach 0 :)(18 votes)

- At3:36, why is the slope zero? isn't the left-sided limit approaching the point c?(7 votes)
- Remember the slope is rise/run – i.e. the "steepness" of a line (or curve). The left sided limit of the
**function**is f(c), but the line is horizontal and therefore has a**derivative**(slope) of zero.(6 votes)

- Is there a reason the derivative at x=a is defined by finding the slope of a secant between the points at x=a and x=a+h where h approaches 0 instead of be defined by finding the slope of a secant between the points at x=a-h and x=a+h as h approaches 0? In other words, why is it:
`f'(x) = lim ( f(x+h) - f(x) ) / ( (x+h) - x )`

h->0

instead of`f'(x) = lim ( f(x+h) - f(x-h) ) / ( (x+h) - (x-h) )`

h->0

If it were the latter, than the derivatives of discontinuous lines and "sharp" points (such as`f(x) = |x|`

at`x=0`

) would be defined. Is there an application where it matters that the derivatives of discontinuous functions or "sharp" points are not defined?(6 votes)- These are actually equivalent forms! Try showing this by applying L'hopital's rule to find both limits. You should find that they both evaluate to 𝑓'(𝑥).(5 votes)

- How about functions like (x²+2x-3)/(x+3)? The function simplifies to {x-1; x≠-3}; does that mean it can't be differentiated at x=-3 even though it can be differentiated into 1 at every other point?(6 votes)
- Correct -- that function can not be differentiated at x=-3, which is a removable discontinuity — i.e. your function is not defined at that point.

Derivatives are only defined at points where the original function is defined — Sal addresses this starting around6:30.(3 votes)

- Is the tangent line to a linear function is the same as the linear function ?(3 votes)
- Yes. The tangent line would have the same slope as the function, and touch the function at some point, and knowing the slope and one point is enough to determine a line.(8 votes)

- Hey folks. Okay So sort of a question that has been prodding at me for a while. I get the reason why you can't find the derivative at a 'sharp point'. But what actually defines a "sharp point" or sharp bend? It isn't very mathematical to just say "hmm this is a sharp point so we can't find the derivative for that" There must be some algebra that tells us when a point is "too sharp" for us to take the derivative right? How do we know this? For example Absolue Value equations produce sharp points, we can't take the derivatives of those I assume. But what else? Surely there must be more times when a graph has a sharp corner that isn't an Absolute Value? So what is it, what values or parts of the function that will tell us okay it's a rounded corner so it's fine...up until.... now! Now it's too sharp to take the derivative..I hope this makes sense..and thanks in advance.(3 votes)
- Being able or unable to take a derivative at that point
*is*what defines a sharp corner. A corner occurs at x=c when a function is differentiable in the neighborhood of c, and not differentiable but still continuous at c.(3 votes)

- How do I develop an equation for a line tangent to the curve at the point defined by a given value?(3 votes)
- There not much too it.

You just need to know how differentiate, and know how to find an equation a straight line.

m(x-x1) = y-y1(3 votes)

- I was just wondering if have some function fx and we find the equation of the tangent line for the derivative of that function at some x, how would it relate to the fx or the derivative of x?(3 votes)
- If 𝑓 '(𝑥) is the derivative of 𝑓(𝑥), then the tangent line to 𝑓(𝑥) at 𝑥 = 𝑎 is the line that passes through the point (𝑎, 𝑓(𝑎)) and has slope 𝑓 '(𝑎)(3 votes)

## Video transcript

- [Instructor] What we're
going to do in this video is explore the notion of
differentiability at a point. And that is just a fancy way of saying does the function have a
defined derivative at a point? So let's just remind ourselves
a definition of a derivative. And there's multiple ways of writing this. For the sake of this
video, I'll write it as the derivative of our function at point C, this is Lagrange notation
with this F prime. The derivative of our function
F at C is going to be equal to the limit as X approaches Z of F of X, minus F of C, over X minus C. And at first when you see this formula, and we've seen it before, it
looks a little bit strange, but all it is is it's
calculating the slope, this is our change in the
value of our function, or you could think of
it as our change in Y, if Y is equal to F of X,
and this is our change in X. And we're just trying to
see, well, what is that slope as X gets closer and closer
to C, as our change in X gets closer and closer to zero? And we talk about that in other videos. So I'm now going to make a
few claims in this video, and I'm not going to
prove them rigorously. There's another video that
will go a little bit more into the proof direction. But this is more to get an intuition. And so the first claim
that I'm going to make is if F is differentiable, at X equals C, at X equals C, then F is continuous at X equals C. So I'm saying if we know
it's differentiable, if we can find this limit, if we can find this
derivative at X equals C, then our function is also
continuous at X equals C. It doesn't necessarily
mean the other way around, and actually we'll look
at a case where it's not necessarily the case the
other way around that if you're continuous, then
you're definitely differentiable. But another way to interpret
what I just wrote down is, if you are not continuous,
then you definitely will not be differentiable. If F not continuous at X equals C, then F is not differentiable, differentiable at X is equal to C. So let me give a few examples
of a non-continuous function and then think about would we
be able to find this limit. So the first is where
you have a discontinuity. Our function is defined at
C, it's equal to this value, but you can see as X
becomes larger than C, it just jumps down and
shifts right over here. So what would happen if you
were trying to find this limit? Well, remember, all this
is is a slope of a line between when X is some arbitrary value, let's say it's out here,
so that would be X, this would be the point X comma F of X, and then this is the point C
comma F of C right over here. So this is C comma F of C. So if you find the left side
of the limit right over here, you're essentially saying
okay, let's find this slope. And then let me get a little bit closer, and let's get X a little bit closer and then let's find this slope. And then let's get X even closer than that and find this slope. And in all of those
cases, it would be zero. The slope is zero. So one way to think about it,
the derivative or this limit as we approach from the left,
seems to be approaching zero. But what about if we were
to take Xs to the right? So instead of our Xs being there, what if we were to take
Xs right over here? Well, for this point, X comma F of X, our slope, if we take F of X minus F of C over X minus C, that would
be the slope of this line. If we get X to be even closer,
let's say right over here, then this would be the slope of this line. If we get even closer,
then this expression would be the slope of this line. And so as we get closer and
closer to X being equal to C, we see that our slope is actually approaching negative infinity. And most importantly, it's approaching a very different value from the right. This expression is approaching
a very different value from the right as it is from the left. And so in this case, this
limit up here won't exist. So we can clearly say this
is not differentiable. So once again, not a proof here. I'm just getting an
intuition for if something isn't continuous, it's pretty
clear, at least in this case, that it's not going to be differentiable. Let's look at another case. Let's look at a case where we
have what's sometimes called a removable discontinuity
or a point discontinuity. So once again, let's say we're
approaching from the left. This is X, this is the
point X comma F of X. Now what's interesting is
where as this expression is the slope of the line
connecting X comma F of X and C comma F of C, which is
this point, not that point, remember we have this removable discontinuity right over here, and so this would be this
expression is calculating the slope of that line. And then if X gets even closer
to C, well, then we're gonna be calculating the slope of that line. If X gets even closer to C,
we're gonna be calculating the slope of that line. And so as we approach from
the left, as X approaches C from the left, we
actually have a situation where this expression
right over here is going to approach negative infinity. And if we approach from
the right, if we approach with Xs larger than C, well,
this is our X comma F of X, so we have a positive slope
and then as we get closer, it gets more positive, more positive approaches positive infinity. But either way, it's not
approaching a finite value. And one side is approaching
positive infinity, and the other side is
approaching negative infinity. This, the limit of this
expression, is not going to exist. So once again, I'm not
doing a rigorous proof here, but try to construct a
discontinuous function where you will be able to find this. It is very, very hard. And you might say, well,
what about the situations where F is not even defined
at C, which for sure you're not gonna be continuous
if F is not defined at C. Well if F is not defined
at C, then this part of the expression
wouldn't even make sense, so you definitely wouldn't
be differentiable. But now let's ask another thing. I've just given you good
arguments for when you're not continuous, you're not
going to be differentiable, but can we make another claim
that if you are continuous, then you definitely
will be differentiable? Well, it turns out that there are for sure many functions, an infinite
number of functions, that can be continuous at
C, but not differentiable. So for example, this could be
an absolute value function. It doesn't have to be an
absolute value function, but this could be Y is equal to the absolute value of X minus C. And why is this one not
differentiable at C? Well, think about what's happening. Think about this expression. Remember, this expression
all it's doing is calculating the slope between the point X comma F of X and the point C comma F of C. So if X is, say, out here,
this is X comma F of X, it's going to be calculated,
so if we take the limit as X approaches C from the left, we'll be looking at this slope. And as we get closer, we'll
be looking at this slope which is actually going to be the same. In this case it would be a negative one. So as X approaches C from the left, this expression would be negative one. But as X approaches C from the right, this expression is going to be one. The slope of the line that
connects these points is one. The slope of the line that
connects these points is one. So the limit of this expression,
or I would say the value of this expression, is
approaching two different values as X approaches C from
the left or the right. From the left, it's
approaching negative one, or it's constantly negative
one and so it's approaching negative one, you could say. And from the right, it's one, and it's approaching one the entire time. And so we know if you're
approaching two different values from on the left side or
the right side of the limit, then this limit will not exist. So here, this is not, not differentiable. And even intuitively, we
think of the derivative as the slope of the tangent line. And you could actually
draw an infinite number of tangent lines here. That's one way to think about it. You could say, well, maybe
this is the tangent line right over there, but
then why can't I make something like this the tangent line? That only intersects at
the point C comma zero. And then you could keep
doing things like that. Why can't that be the tangent line? And you could go on and on and on. So the big takeaways here,
at least intuitively, in a future video I'm
going to prove to you that if F is differentiable at
C that it is continuous at C, which can also be interpreted
as that if you're not continuous at C, then you're
not gonna be differentiable. These two examples will hopefully give you some intuition for that. But it's not the case that
if something is continuous that it has to be differentiable. It oftentimes will be
differentiable, but it doesn't have to be differentiable, and
this absolute value function is an example of a
continuous function at C, but it is not differentiable at C.