Lesson Video: The Differentiability of a Function | Nagwa Lesson Video: The Differentiability of a Function | Nagwa

# Lesson Video: The Differentiability of a Function Mathematics • Second Year of Secondary School

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In this video, we will learn how to determine whether a function is differentiable and identify the relation between a functionβs differentiability and its continuity.

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### Video Transcript

The Differentiability of a Function

In this video, weβll learn how to determine whether a function is differentiable. And weβll identify the relationship between a functionβs 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 weβll mainly be using the first one.

Now, a key point that weβll 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 Leibnizβs 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 Lagrangeβs notation. Both of these notations are very commonly used. And youβll 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 weβll 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, weβve 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, itβs 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 itβs worth noting that we could have made a more rigorous argument based on the limits which define a derivative. However, this example illustrates that itβs 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, weβve sketched out a number of graphs and weβll 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. Letβs 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, itβs 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. Weβve now successfully applied the power rule. Snd weβve 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 itβll 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 weβve 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. Letβs 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 whatβs 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 theyβre 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 weβre 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 weβve 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, weβll 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, weβre 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. Weβve 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, weβll 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, weβre 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, itβs 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 itβs 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. Letβs 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, weβll 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, weβve 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. Letβs 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 weβll 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 thatβs, in fact, what weβre 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₯. Weβll 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. Weβve 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, weβve illustrated the process for evaluating whether a function is differentiable at a point. Weβve done this, utilizing the relationship between differentiability and continuity alongside the tools of differentiation, such as the power rule. To summarize, Letβs go through a few key points. The derivative can be represented using either Leibnizβs notation or Lagrangeβs 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.

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