If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

### Course: Class 12 math (India)>Unit 5

Lesson 5: Differentiability

# Differentiability at a point: algebraic (function is differentiable)

We examine a piecewise function to determine its continuity and differentiability at an edge point. By analyzing left and right hand limits, we establish continuity. Checking the limit of the difference quotient confirms both left and right hand limits are equal, making the function continuous and differentiable at the edge point.

## Want to join the conversation?

• Why isn't this an example of the non-differentiable "sharp turn" discussed in the Differentiability at a Point: Graphical video?
• Not all piecewise functions are sharp turns. In this case, if we compute the limit that defines the derivative, from both sides of 3, we find that the two limits are equal. This means that the slope of the tangent is equal from both sides, so the derivative exists.
• I feel like, even though I've watched all videos in this section up to this point, some of this came way out of left field. In which video can we find the difference between x---> 0, h---> 0, x--->3, etc? This video also seems to be talking about two different types of limits (limits to what y can equal, limits to what the slope can equal?). Where did this come from?
• Yes, two different limits are mentioned in the video. One is to check the continuity of f(x) at x=3, and the other is to check whether f(x) is differentiable there.

First, check that at x=3, f(x) is continuous. It's easy to see that the limit from the left and right sides are both equal to 9, and f(3) = 9.

Next, consider differentiability at x=3. This means checking that the limit from the left and right sides of f'(x) are both equal. In this case, they are both equal to 6. Therefore f(x) is differentiable at x=3.
• At , can u describe the "equation" for the Differentiability? Why do u write (x-3) under there?
• How can we know that this is not a "sharp turn"? This is similar to the top comment but I don't think the replies really answer this question.
• A sharp turn can be visualized by imagining the tangent line of either side of the "apex" of the turn (i.e. the tip/point of the sharp turn).
If you imagine the tangent to the "left" and "right" of that point, you'll see the tangent lines don't smoothly touch each other. Imagine a smooth hill now, of a function that CAN be differentiated, if you imagine those tangent lines as they get closer to the top of the hill they begin to "line up" with each other, and finally meet horizontally!

The real way to check this is to see what the slope of each tangent line is from either side. Normally we take the limit x-> c+ and limit x->c- (to represent approaching from the right and left respectively). If these limits are not equal, then we are getting 2 different slopes at a single point for the tangent line! Obviously this isn't what we want, which slope/tangent line would we choose?! Which limit would we choose? We cannot definitively say that it is due to a sharp turn, but it doesn't matter, this method works whether there's a sharp turn or other discontinuity! The way to tell for sure would be to see if the function is continuous. If it is continuous function (like f(x) = |x|) AND the limits don't work from either side (like f(x) = |x|), then it's a sharp turn! Can you prove/see why?
• It is not fully clear where the differentiability requirement is coming from
• I'm not quite sure what you mean by the differentiability requirement. I'm guessing you're asking how Sal knows that it's differentiable at that point.

A piecewise function is differentiable at a point if both of the pieces have derivatives at that point, and the derivatives are equal at that point. In this case, Sal took the derivatives of each piece: first he took the derivative of x^2 at x=3 and saw that the derivative there is 6. Then, he took the derivative of 6x-9 at x=3, and saw that the derivative there is also 6.

It might be more clear after you watch the next video, where he shows an example that is not differentiable.
• Wait, he gets 9 from the same formula he then subtracts 9 from later, why is this not just turning into (9-9)/(3-3) over and over?
• That's why he factors the enumerator and cancels the `(x-3)`! `(9-9)/(3-3) = undefined`, and you can't have that. So, by cancelling factors, Sal is able to sidestep the undefined values and finish solving the equation.
(This is an example of a removable discontinuity. There is a hole when `x = 3`, but the limit at `x = 3` exists, so you can still take the derivative.)
I hope this helps!
• I understand that this is supposed to be algebraic way to find differentiability at a point, but I wanted to share an easier way that involves using calculus to find if it is differentiable or continuous at a certian point.

Using the example in the video:
Is the function continuous/differentiable at x=3?
f(x)=
|x^2,x<3(1)
|6x-9,x≥3(2)
Firstly, we substitute x=3 into both equations to find if they are continuous.
(1): f(x)=9(from 3^2)
(2): f(x)=9(from 18-9)
Because both f(x) values are the same, we know that the function is continuous at x=3(if they arent the same, then the function is not continuous).

To find if they are differentiable, we must take the derivative of both equations.
f'(x)=2x (1)
f'(x)=6 (2)
If we again substitute x=3, we get 6 in both. Therefore, we know that the function is differentiable.
• Do we actually need to test for continuity? If we find that the function is differentiable at x=3, doesn't it automatically follow that it is continuous at x=3?
• a) Yes, you actually need to test for continuity as not all continuous functions are derivable.

b) Yes, derivability implies continuity but not the other way around.
• I have done examples and practice problems where you just differentiate the function as it is, without thinking about either side. Why is this different? Is it because it because it might not be continous? but we already see the it is. Why have I done practice problems where I don't check for differentiability from both side, and now I am doing it from both sides. Is this method just better practice?
• differentiation is defined from the first principles of differentiation which involves a limit.

So you need sure that limit exists to ensure the function is differentiable.

The function also needs to be continuous for it to be differentiable because of the definition.