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## APÂŪïļ Calculus AB (2017 edition)

### Course: APÂŪïļ Calculus AB (2017 edition)Â >Â Unit 2

Lesson 5: Differentiability

# 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.
• 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......
• At , why is the expression approaching negative infinity?
• 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.....
• 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?
• 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 :)
• At , why is the slope zero? isn't the left-sided limit approaching the point c?
• Remember the slope is rise/run â 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.
• 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``

``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?
• 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 ð'(ðĨ).
• How about functions like (xÂē+2x-3)/(x+3)? The function simplifies to {x-1; xâ -3}; does that mean it can't be differentiated at x=-3 even though it can be differentiated into 1 at every other point?
• Correct -- that function can not be differentiated at x=-3, which is a removable discontinuity â i.e. your function is not defined at that point.

Derivatives are only defined at points where the original function is defined â Sal addresses this starting around .
• Is the tangent line to a linear function is the same as the linear function ?
• 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.
• 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.
• 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.
• How do I develop an equation for a line tangent to the curve at the point defined by a given value?
• 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