This is a note written to address an issue we had in Tueday’s tutorial. Specifically the method I have shown you for doing one of the types of differentiability questions is somewhat flawed. There are a number of further assumptions that we need to make in order to make the analysis correct. I apologise for this oversight. However all is not lost – we can still ‘fix’ our (easier) method by taking these additional assumptions into account.

**This question will now not be examinable in your summer exam.**

**The Question**

*Let be defined by:*

*Show that is differentiable but not twice differentiable.*

**The Flawed* Solution**

Away from , is a polynomial and hence differentiable with derivative:

Now

.

and

.

Hence exists as the left and right derivatives are equal, hence is differentiable.

Away from , is a polynomial and hence twice differentiable:

Now

and

Hence does not exist hence is *not* twice differentiable

*** ***this method leads to the correct conclusion, and it will do so for all cases we would encounter in MS 2001.*

*Essentially the error that was made by me was confusion in the definition of the left- and right-hand derivatives. Let and .*

*Left- and right-hand Derivatives: Correct Definition*

Right-hand Derivative at : .

Left-hand Derivative at : .

*Left- and right-hand Derivatives: Incorrect Definition*

Right-hand Derivative at : .

Left-hand Derivative at : .

**Theorem**

*Let and . is differentiable at if and only if the left- and right-hand derivatives exist and are equal.*

**Towards a Correct Analysis**

* The above flawed solution more-or-less looks at the function and says well it’s smooth there because the derivatives are *matching up on both sides – *i.e as a function, is continuous so must be differentiable (click to sharpen):*

*A plot of .*

*We will see later that by this flawed analysis a discontinuous function can be differentiable with continuous derivative (!).*

**The Correct Solution**

Away from , is a polynomial and hence differentiable with derivative:

Now

.

and

.

Hence exists as the left- and right-hand derivatives are equal, hence is differentiable.

Away from , is a polynomial and hence twice differentiable:

Now

and

Hence does not exist hence is *not* twice differentiable

**What is wrong with the “flawed solution”?**

Two counterexamples show that our original analysis is flawed (the second is not as important to us but is relevant in a subsequent proof):

**Example 1:**

Let

*A plot of *

Now geometrically it is clear that is does not have a well-defined tangent at . Even more straightforwardly, we can see that is not continuous at and recalling the following theorem (and it’s contrapositive):

differentiable continuous

not continuous not differentiable,

It is clear that is not differentiable at yet

thus,

and so our flawed analysis would suggest that is differentiable at (and indeed infinitely differentiable). Our analysis is suggesting that the discontinuous function is differentiable with continuous derivative (!).

Hence if the ‘flawed method’ is to work it will be necessary to show that the function is continuous.

**Example 2**

Let

Now away from , has derivative but at the tangent is vertical – hence the slope of the tangent is undefined ( undefined). Indeed plugging in we get:

Uh-oh! However

which could lead to the nonsensical conclusion that is differentiable at with derivative .

Hence if the ‘flawed method’ is to work it will be necessary to show that the derivative is *bounded *(i.e. not infinite).

**Theorem**

*Suppose that is continuous on and differentiable except perhaps at . Suppose further that ** exists. Then is differentiable at with derivative *.

*Proof: *Let . is continuous on and differentiable on . Then, by the Mean Value Theorem, such that

Now taking the limit as on both sides:

Now as hence the left hand side is equal to :

Similarly taking and we can show:

Hence is differentiable at with derivative

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