### Video Transcript

The Differentiability of a
Function

In this video, weβll learn how to
determine whether a function is differentiable. And weβll identify the relationship
between a functionβs differentiability and its continuity. The process of finding a derivative
is called differentiation. Since derivatives are one of the
fundamental building blocks of calculus, differentiation is, therefore, a very
important tool. It, therefore, follows that being
able to tell whether a function is differentiable or not can be very useful for
us. We recall that the derivative
measures the rate at which the output value of a function π changes with respect to
a change in its input π₯.

In a more formal since the
derivative can be defined using limits. The derivative of a function π at
the point where π₯ is equal to π₯ nought is defined as the limit as β approaches
zero of π of π₯ nought plus β minus π of π₯ nought all divided by β. Now it may help you to think of the
bottom half of this quotient as π₯ nought plus β minus π₯ nought, which is, of
course, just β. In this way, we see that our
formula looks a lot like the change in π¦ divided by the change in π₯ as our changes
become infinitesimally small. Now, an equivalent definition for
the derivative would be the limit as π₯ approaches π₯ nought of π of π₯ minus π of
π₯ nought all divided by π₯ minus π₯ nought. Both of these definitions are
commonly used, but for the purpose of this video weβll mainly be using the first
one.

Now, a key point that weβll be
revisiting throughout this video is that the derivative only exists if the limit
which defines it exists. If the limit and therefore the
derivative do exists at a point, we say that at this point our function is
differentiable. Before we continue, we should know
that there are two different forms of notation for our derivatives. If we say that π¦ is equal to our
function π of π₯, then the derivative can be written as follows. The first way is dπ¦ by dπ₯, which
is so-called Leibnizβs notation. This notation uses the
infinitesimals that we see here dπ¦ and dπ₯.

The second way is π dash of π₯ or
π prime of π₯. And this is so-called Lagrangeβs
notation. Both of these notations are very
commonly used. And youβll be seeing them in this
video. Returning to our theory, if we
imagine π of π₯ is a curve, the derivative of π of π₯ would then represent the
tangent to that curve. It would then make sense that if we
cannot define the tangent of our curve, then the derivative does not exist. Observing the graph of a function
can often give us a visual understanding of when the function is and is not
differentiable, as weβll see in the following example.

The figure shows the graph of
π. What can be said about the
differentiability of π at π₯ equals negative four?

Here, weβve been given a graph
which is defined over the interval from negative seven to negative one. Through this interval, we see that
our curve is smooth at all points, aside from the point where π₯ is equal to
negative four. At this point of coordinates
negative four, five, we have a sharp corner. This means that the slope of the
tangent, just to the left of π₯ equals negative four, will be different to the slope
of the tangent just to the right of π₯ equals negative four. Here, we can even go so far as to
say that one of our slopes will be positive and one of our slopes will be
negative. Given that we have two different
tangents on either side, it follows that it is not possible to define a tangent at
π₯ equals negative four. And therefore, itβs also not
possible to define the derivative.

If we were to imagine the graph of
π¦ equals π dash of π₯ our first derivative, we would expect to see a sharp jump in
the π¦-value when π₯ equals negative four. From our observations, we conclude
that the function is not differentiable at π₯ equals negative four because the
functions rate of change is different on both sides of that point. And with this statement, we have
answered our question.

Now itβs worth noting that we could
have made a more rigorous argument based on the limits which define a
derivative. However, this example illustrates
that itβs sometimes possible to quickly evaluate the differentiability of a function
based on its graph.

Alongside the sharp corner that we
saw in the previous example, there are many different reasons that a function π may
not be differentiable at some point where π₯ is equal to π₯ nought. All of these cases can be
understood by recalling our definition of the derivative and considering when this
limit can be said to exist or not. Here, weβve sketched out a number
of graphs and weβll go through each of their implications in turn. In the first two cases, we have a
discontinuity and a corner. In these cases the left- and the
right-sided the limits as β approaches zero would disagree. Since they disagree, this also
tells us that the normal limit that defines the derivative does not exist. And hence, the functions would not
be differentiable at π₯ nought.

For the next two cases, we have a
cusp and a vertical tangent. We know that, on a graph, vertical
lines or lines that tend to the vertical correspond to gradients of positive or
negative infinity. This means that our left- and our
right-sided limits as β approaches zero would also take values of positive or
negative infinity. In the case of a cusp, these left-
and right-sided limits would disagree, taking values of infinity with opposite
signs, whereas in the case of a vertical tangent, they would agree. We already know the story for
disagreeing one-sided limits. However, even if these were to
agree, for example, both taking the value of infinity, and we were to say that our
normal limit is also equal to positive infinity. This is still a particular way of
expressing that the limit does not exist, since infinity is not a number but rather
a concept.

Finally, in the case of oscillating
behaviour, we observe more and more frequent oscillations as our value of π₯
approaches π₯ nought. It should also then be clear that
the slope of our graph representing the derivative is also oscillating more and more
rapidly as our value of π₯ approaches π₯ nought or as β approaches zero. This means that it does not make
sense for us to assign a value to our limit as β approaches zero. And we, therefore, say that this
limit does not exist. As for all of these cases, the
limit does not exist. And therefore, the derivative does
not exist. And we can, therefore, say that our
function is not differentiable at π₯ nought.

Moving on, alongside evaluating a
limit, there are a number of different other tools that can help us when
differentiating a function. One such example of this is the
power rule, which tells us that if some function π of π₯ takes the form π₯ to the
π, then the derivative of this function would be π times π₯ to the power of π
minus one. We note here that our steps have
been to multiply by the power for which our π₯ has been raised to. And we then reduce this power by
one. Letβs take a look at one example
which utilizes the power rule.

Consider the function π of π₯ is
equal to the cube root of π₯. Part a), what is the domain of
π?

For part a), we can immediately
recall that the cube root of any real number is well-defined on the real
numbers. If instead we had a square root, we
know that this statement would not be true, since the square root of negative
numbers are not defined in the real numbers. As it stands, however, we can
answer part a) in a very straightforward manner by saying that the domain of π is
β, the real numbers.

Moving on to part b), finding the
derivative of π. One of the tools that we have at
our disposal is the power rule. This rule tells us that for some
function π of π₯, which takes the form π₯ to the power of π, the derivative of our
function would be π times π₯ to the power of π minus one. To apply this to our question, itβs
helpful to express the cube root of π₯ as π₯ to the power of one over three. We can then apply the power rule,
multiplying our π₯ by one over three, which is our power and subtracting one from
the power which gives us negative two over three. An equivalent way to express this
would be one over three times the cube root of π₯ squared. Weβve now successfully applied the
power rule. Snd weβve sold part b) finding an
expression for the derivative of our function π.

Finally, for part c), finding the
domain of this derivative, we used the expression that we just found. For this part of the question, we
must consider all points for which π dash of π₯ is undefined. Since we have π dash of π₯ in the
form of a quotient, we can say that itβll be undefined when the denominator of this
quotient is equal to zero. We, therefore, need to find a value
for π₯, for which three times the cube root of π₯ squared is equal to zero. And the only value which satisfies
this is when π₯ is equal to zero. Now, since π₯ equals zero is the
only point for which π dash of π₯ is undefined over the real numbers, we can say
the following. The domain of the derivative of our
function π dash is the real numbers β minus the set which contains zero.

We have now solved all three parts
of our question.

And we can take note of the fact
that the domain of some function π is not necessarily the same as the domain of its
derivative. In the example that weβve just
seen, instead of evaluating a limit to find our derivative, we instead used the
power rule to speed up the process. There are, however, some examples
where the power rule might not give us a complete understanding of our function. Letβs take a look at one such
example to illustrate this.

Suppose π of π₯ is equal to
negative six π₯ minus four, if π₯ is less than or equal to negative one, and three
π₯ squared if π₯ is greater than negative one. What can be said about the
differentiability of the function at π₯ equals negative one?

Here, we have been given a
piecewise defined function composed of two subfunctions, a binomial and a
monomial. Both of these subfunctions on their
own or whatβs known as smooth and will be defined over all the real numbers. In fact, on their own, we can say
that all polynomials are smooth, which means that theyβre differentiable on the real
numbers. For a piecewise defined function,
however, we must check the point at which are two subfunctions join. That is, in this case, π₯ equals
negative one. To begin the process, let us
differentiate both of our subfunctions using the power rule.

The derivative of negative six π₯
minus four is negative six times π₯ to the power of zero, which is, of course, just
negative six. The derivative of three π₯ squared
is two times three π₯, which is, of course, six π₯. We can represent this more
succinctly by saying that π dash of π₯ is equal to a negative six, if π₯ is less
than negative one, and six π₯ if π₯ is greater than negative one. It should be noted that even though
we see a less than or equal to inequality symbol here, we are not yet making any
claims as to the value of our derivative when π₯ is equal to negative one, since
this is what weβre trying to find at the moment.

Considering the derivative either
side of the point where our two subfunctions join, we might try to move forward by
substituting the value of π₯ equals negative one into the two subfunctions that we
just found for π dash of π₯. Doing so, we would find that both
of our values are negative six. We may then conclude that since
these two values are equal, that our function is differentiable at the point where
π₯ is equal to negative one. Unfortunately, this is not the
case. And to see why. Let us recall the definition of the
derivative as a limit.

Here weβve shown the limit which
defines the derivative. To move forward, we recognize that
since our function is defined differently on either side of the point where π₯
equals negative one, the left- and the right-sided limits will, therefore, have
different expressions. Remembering that our π₯ nought is
equal to negative one, we can begin by expressing the left limit as follows. Now, clearly, we cannot simply
substitute β equals zero into this expression. Otherwise, weβll be left with the
indeterminate form of zero over zero. Instead, we know that when π₯ is
less than negative one, π of π₯ is equal to negative six π₯ minus four. Hence π of negative one plus β
will evaluate to two minus six β.

By similar reasoning, when π₯ is
equal to negative one π of π₯ takes the same subfunction. And therefore, π of negative one
evaluates to two substituting these in we find that the expression for our
left-sided limit becomes two minus six β minus two all over β, which simplifies to
negative six β over β. Now at this stage, since we know
that β is approaching zero and not equal to zero, weβre allowed to cancel the common
factor on the top and bottom half of our quotient. We are then left with the limit as
β approaches zero from the left of negative six which, of course, is just equal to
negative six.

Now for the right-sided limit, by
similar reasoning as previously, we know that when π₯ is greater than negative one,
π of π₯ is defined by the subfunction three π₯ squared. Hence, in this case, π of negative
one plus β evaluates to three minus six β plus three β squared. For π of negative one, we should
be careful not to use our three π₯ squared again. Since when π₯ is equal to negative
one, π is defined by our other subfunction. Weβve already found the value of
this to be two earlier. Again, we go through the same
process of substituting these in and simplifying the expression for our limit. This time we reach a different
result. Looking at the first term of our
limit, weβll have the limit as β approaches zero from the positive direction of one
over β, which is equal to infinity. By extension, this means our entire
right-sided limit is also equal to infinity.

We know that this is a particular
way of expressing that the limit does not exist. If the right limit does not exist,
then this also tells us that the normal limit does not exist. And hence, the derivative is not
defind. The reason this is the case is that
our graph actually has a discontinuity at π₯ equals negative one. And if we were to draw it, we would
see this. Since we have concluded that the
derivative is undefined at π₯ equals negative one, weβre also in a position to say
that the function is not differentiable at π₯ equals negative one. And in fact, this is the answer to
our question.

As we saw in our example, to check
the differentiability of a function at a point, it is not necessarily enough to
check that the left- and the right-sided limits of the derivative of the function
agree at that point. Instead, for some value π₯ nought,
we must also check that the left- and the right-sided limits of the function itself
as π₯ approaches π₯ nought, both exist, agree, and are equal to the value of the
function evaluated at π₯ nought. Perhaps a more familiar form of
this can be obtained by recalling that if these two conditions are satisfied, then
the normal limit also exists and takes the same value. In fact, this is the condition for
continuity.

An important general rule that we
can use is if a function is differentiable at a point π₯ nought, then it is also
continuous at that point. For this rule, a logically
equivalent statement would be if the function is not continuous at a point π₯
nought, itβs also not differentiable at π₯ nought. Now, we must be a little careful
not to overextend this rule and end up with a false conclusion that if a function is
continuous at π₯ nought, it is also differentiable at π₯ nought. This is in fact, false. And the following illustration may
help us understand this.

Consider that our orange circle
represents all functions which are continuous. Our pink circle then represents all
functions which are differentiable. As we can see, our differentiable
circle is contained within the circle of all functions which are continuous. If we had a function π of π₯ which
is inside our differentiable circle, we would then know for sure it is also inside
our continuous circle. Now, consider some other function
π of π₯, which is not inside our continuous circle. Looking at this, we know for sure
it is also not inside our differentiable circle.

Now, however, if we consider some
third function β of π₯, it is possible for this to be inside our continuous circle
without necessarily being inside our differentiable circle. This means that if β of π₯ is
continuous, it may or may not be differentiable. And we cannot draw conclusions just
based on its continuity. In fact, there are many examples of
functions which are continuous but not differentiable. And we have already seen this in
the corner that we evaluated for our first question. When evaluating whether a function
is differentiable at π₯ nought, we can hence go through the following steps.

We first check that our function π
is continuous at π₯ nought. If itβs not, we conclude that our
function is not differentiable at π₯ nought. However, if it is, we then check
that the left- and the right-sided limits as π₯ approaches π₯ nought of our
derivative π dash of π₯ exists and agree. Again, if not, we conclude our
function is not differentiable at π₯ nought. However, if this is the case. Then we conclude that our function
π is differentiable at π₯ nought. Letβs go through one final example
to illustrate this process.

Suppose that π of π₯ is equal to
negative one plus three over π₯ if π₯ is less than or equal to one and negative π₯
cubed plus three if π₯ is greater than one. What can be said about the
differentiability of π at π₯ equals one?

Here, we have been given a
piecewise defined function and asked to check the differentiability. If we call our value of one π₯
nought, our general process for this type of question is first to check for
continuity at π₯ nought. And then if this is satisfied, we
check that the limit as π₯ approaches π₯ nought of our derivative exists, which
means checking the left and right limits both exists and agree. We first begin with the continuity
condition stated here. Since this is a piecewise defined
function, weβll need to check that both the left and the right limits as π₯
approaches one of π of π₯ both exist and agree. When π₯ is less than one, our
function is negative one plus three over π₯. By direct substitution, our
left-sided limit then evaluates to two. When π₯ is greater than one, our
function is negative π₯ cubed plus three. And by the same approach, the
right-sided limit also evaluates to two.

Since these two limits both exist
and agree, we can also see that the normal limit exists and is equal to two. Moving onto π of one, the function
itself here is defined by our first subfunction since π₯ is equal to one. And in fact, weβve already
substituted one into our first subfunction here, which gave us an answer of two. This means that we can say that the
limit as π₯ approaches one of π of π₯ and π of one are both equal to two. We have, therefore, satisfied the
continuity condition and our function is continuous when π₯ is equal to one. Letβs now move on to considering
the derivative of our function.

Since our function is defind
peacewise, the derivative of our function will also be defined piecewise. And weβll need to differentiate
both of our subfunctions. And here is worth noting that were
not yet making any claims as the derivative when π₯ is equal to one, since thatβs,
in fact, what weβre trying to find right now. One of the tools that we can use to
help us differentiate our subfunctions is the power rules stated here. To help us apply this rule more
easily, we can reexpress three over π₯ as three π₯ to the power of minus one. Applying the rule, we find that π
dash of π₯ is equal to negative three over π₯ squared if π₯ is less than one and
negative three π₯ squared if π₯ is greater than one.

We now need to consider the limit
as π₯ approaches one of π dash of π₯. Weβll do so by considering the
left- and the right-sided limits in a process similar to the one that we just
reformed for the continuity condition. By direct substitution of one, we
find that both of these limits are equal to negative three. And since they exist and agree, the
limit as π₯ approaches one of π dash of π₯ is also equal to negative three. Since this limit exists, we know
that our derivative also exists at π₯ equals one. Weβve now completed all the steps
of our process. And from these, we can conclude
that the function π is differentiable at the point where π₯ is equal to one.

In this example, weβve illustrated
the process for evaluating whether a function is differentiable at a point. Weβve done this, utilizing the
relationship between differentiability and continuity alongside the tools of
differentiation, such as the power rule. To summarize, Letβs go through a
few key points. The derivative can be represented
using either Leibnizβs notation or Lagrangeβs notation. And we have two alternative, but
equivalent, definitions for this, both of which utilize limits. When the limit does not exist, the
function itself is not differentiable at this point. And there are a number of different
ways this can occur. And finally, we can draw
conclusions about our function based on the relationship between differentiability
and continuity.