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Differentiability and continuity

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

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  • aqualine seed style avatar for user Jason  Li
    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)
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    • duskpin ultimate style avatar for user Davide Ghazal
      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......
      (39 votes)
  • marcimus orange style avatar for user Grace Pfohl
    At , why is the expression approaching negative infinity?
    (18 votes)
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    • duskpin ultimate style avatar for user Davide Ghazal
      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.....
      (21 votes)
  • starky sapling style avatar for user 20leunge
    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)
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    • aqualine ultimate style avatar for user stolenunder
      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)
  • duskpin seedling style avatar for user Latifa
    At , why is the slope zero? isn't the left-sided limit approaching the point c?
    (7 votes)
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  • piceratops ultimate style avatar for user Khan Gressman
    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)
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  • piceratops ultimate style avatar for user Richard
    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)
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    • female robot grace style avatar for user tyersome
      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 around .
      (3 votes)
  • male robot johnny style avatar for user Mohamed Ibrahim
    Is the tangent line to a linear function is the same as the linear function ?
    (3 votes)
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  • primosaur ultimate style avatar for user Steve
    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)
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    • leaf green style avatar for user kubleeka
      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)
  • duskpin seedling style avatar for user Andrew Carpenter
    How do I develop an equation for a line tangent to the curve at the point defined by a given value?
    (3 votes)
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  • spunky sam blue style avatar for user Paramvir Singh
    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)
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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.