### Video Transcript

Definition of The Derivative

In this video, we will learn how to calculate the derivative of a function using the formal definition of its derivative as a limit. Weβll be covering the definition of a derivative in greater detail and then looking at some examples. Letβs start by exploring the definition of a derivative at a particular point.

We know that the derivative of some function π of π₯ tells us about the rate of change of that function at a particular point. Letβs choose that point to be π₯ nought. And this could be any point. And we know that the rate of change of the function at this point is equal to the slope of the function at this point. Therefore, we can find an estimate for the derivative by approximating the tangent to the curve at this point and then finding the slope of this tangent. Letβs now consider a new graph of a different π of π₯.

If weβre trying to find the derivative of π of π₯ at π₯ nought, then we could estimate it by approximating a tangent at π₯ nought. One way in which we can draw an approximate tangent is to pick another π₯-value which is close to π₯ nought, say π₯ nought plus β where β is a constant. Then, we can draw our approximate tangent as the line passing through π₯ nought, π of π₯ nought and π₯ nought plus β, π of π₯ nought plus β. This is what our approximate tangent would look like. However, this is not a very good approximation.

One way we could improve our approximation for the tangent at this point is to decrease the value of β. Letβs say that we decreased β such that π₯ nought plus β lies at this point on our π₯-axis, then the point π₯ nought plus β, π of π₯ nought plus β, would have shifted to here. Then, our new approximation for the tangent will pass through this point and the point π₯ nought, π of π₯ nought. Therefore, it will look something like this, which is a better approximation for the tangent of π of π₯ at π₯ nought.

We could improve the approximation even more if we decrease the value of β again such that π₯ nought plus β lied at this point on our π₯-axis. This would again shift the point π₯ nought plus β, π of π₯ nought plus β. And we could then draw a new approximation for the tangent at π₯ nought, which we can see is even closer to the actual tangent at π₯ nought. Now, what this method for approximating the tangent at π₯ nought is telling us is that the smaller the value of β, the closer our approximate tangent gets to the real tangent at π₯ nought.

And so, we can say that the limit, as β tends to zero, of our approximate tangent is in fact equivalent to the real tangent of π of π₯ at π₯ is equal to π₯ nought. Now, if we recall what a derivative is, weβll remember that itβs the value of the slope of the function at that particular point. Since the slope of the tangent at any point on our curve represents the slope of the function at that point, we can use the slope of our approximated tangents in order to help us define a derivative.

Letβs recall what the points were which we were using in order to approximate our tangents. We used π₯ nought, π of π₯ nought, and π₯ nought plus β, π of π₯ nought plus β. We know that the slope of any straight line is the change in π¦ over the change in π₯. Therefore, the slope of any of our tangents can be given by π of π₯ nought plus β minus π of π₯ nought over π₯ nought plus β minus π₯ nought. In the denominator, we have π₯ nought plus β minus π₯ nought. So, the two π₯ noughts cancel out, leaving us with the slope of our tangent being equal to π of π₯ nought plus β minus π of π₯ nought all over β.

We will combine the slope of the tangent along with the fact that as π₯ nought gets closer to zero, our approximation of that tangent gets closer to the real tangent at π₯ nought in order to define a derivative. And so, we arrive at the definition of the derivative. We say that the derivative of a function π of π₯, at a point π₯ nought, is defined as the limit as β goes to nought of π of π₯ nought plus β minus π of π₯ nought all over β if the limit exists.

If we allow π₯ one to be equal to π₯ nought plus β, then we have that β is equal to π₯ one minus π₯ nought. And as β goes to zero, π₯ one minus π₯ nought goes to zero, which means that π₯ one tends to π₯ nought. So, we can rewrite this limit in terms of π₯ one and π₯ nought instead of π₯ nought and β. This equivalent definition is the limit as π₯ one tends to π₯ nought of π of π₯ one minus π of π₯ nought over π₯ one minus π₯ nought. And this is again only if the limit exists.

There are a few ways in which we can denote a derivative. The first is using prime notation. This is written as π prime of π₯. We can say that itβs the derivative of π with respect to π₯. Another way we can denote a derivative is using Leibnizβs notation. Which tells us that if we write π of π₯ as π¦, then the derivative is dπ¦ by dπ₯, which also means the derivative of π¦ with respect to π₯. Both of the definitions and notations here will be very useful for us while studying calculus. And so, itβs important for us to be comfortable using both of the definitions and both sets of notation. Letβs now look at some examples of how we can use the definition to find derivatives.

Find the derivative of π of π₯ which is equal to π₯ squared at the point π₯ is equal to two from first principles.

What this question means when it says from first principles is to use the definition of a derivative. Which tells us that the derivative of π with respect to π₯ at π₯ nought is equal to the limit as π₯ one tends to π₯ nought of π of π₯ one minus π of π₯ nought over π₯ one minus π₯ nought. From the question, we can see that π of π₯ is equal to π₯ squared. And weβve been asked to find the derivative at the point π₯ is equal to two. Therefore, π₯ nought is equal to two.

From this, we can say that π prime of two is equal to the limit as π₯ one tends to two of π₯ one squared minus two squared over π₯ one minus two. We can rewrite two squared as four. Now, we see that in the numerator of our fraction we have π₯ one squared minus four, which is the difference of two squares. And so, it can be factorized to π₯ one minus two multiplied by π₯ one plus two.

Here, we notice we have a factor of π₯ one minus two in both the numerator and denominator. And since π₯ one minus two is not equal to zero, we can cancel these here. And so, we find that π prime of two is equal to the limit as π₯ one tends to two of π₯ one plus two. Using direct substitution, we obtain that this is equal to two plus two. And so, we reach our solution, which is that the derivative of π₯ squared at the point π₯ is equal to two is four.

Now, if we remember back at earlier in this video, when we were defining the derivative, we used the slopes of approximations of tangents to get closer and closer to the derivative of the function. In the next example, weβll see how we can use the definition of a derivative to accurately find the slope of a tangent to a function at a point.

Let π of π₯ be equal to eight π₯ squared minus six π₯ plus nine. Use the definition of the derivative to determine π prime of π₯. What is the slope of the tangent to its graph at one, two?

The first thing we need to do is to use the definition of the derivative to determine π prime of π₯. We can recall the definition for the derivative, which tells us that π prime of π₯ is equal to the limit as β goes to zero of π of π₯ plus β minus π of π₯ all over β. Now, in our case, π of π₯ is equal to eight π₯ squared minus six π₯ plus nine, so we can substitute this into our limit. We obtain that π prime of π₯ is equal to the limit as β tends to zero of eight multiplied by π₯ plus β squared minus six multiplied by π₯ plus β plus nine minus eight π₯ squared minus six π₯ plus nine all over β.

Our next step is to expand the brackets. We obtain this limit here. We notice there are a few things we can cancel out in the numerator of our fraction. We have an eight π₯ squared and a negative eight π₯ squared. We have a negative six π₯ and a six π₯. And we also have a nine and a negative nine. Removing these six terms, we are left with the limit as β tends to zero of 16βπ₯ plus eight β squared minus six β all over β.

And we notice we have a factor of β in both the numerator and denominator. Since β is not equal to zero, we can cancel it out. Giving us that π prime of π₯ is equal to the limit as β tends zero of 16π₯ plus eight β minus six. Applying direct substitution to this limit, we reach the solution to the first part of the question, which is that π prime of π₯ is equal to 16π₯ minus six. We can now move on to the second part of the question, which is to find the slope of the tangent to its graph at one, two.

Now, we know that the derivative of π will tell us the slope at any point on π. Therefore, the slope at one, two will be the value of π prime of π₯ when π₯ is equal to one. So, we find π prime of one. This gives us 16 multiplied by one minus six, which simplifies to give us a slope of 10. In the next example, weβll see how we can use the definition of a derivative in order to find the derivative of a reciprocal function.

Using the definition of the derivative, evaluate d by dπ₯ of one over one plus π₯.

We can recall the definition of the derivative. And that is that if we set π¦ to be equal to π of π₯, then dπ¦ by dπ₯ is equal to the limit as β tends to zero of π of π₯ plus β minus π of π₯ all over β. Now, in our case, π¦ is equal to one over one plus π₯. Therefore, we can say that dπ¦ by dπ₯ is equal to the limit as β tends to zero of one over one plus π₯ plus β minus one over one plus π₯ all over β.

We need to express our fraction as a single fraction over a common denominator. And we can do this by finding a common denominator for the two fractions in the numerator. That common denominator is one plus π₯ multiplied by one plus π₯ plus β, leaving us with this limit. And we can combine the two fractions in the numerator, leaving us with the limit as β tends to zero of negative β over β multiplied by one plus π₯ multiplied by one plus π₯ plus β. Since β is not equal to zero, we can cancel β in the numerator and denominator.

Next, we can apply limit rules. We have that the limit of a quotient is equal to the quotient of the limit. Since negative one is a constant, we are left with this. And we can simply apply direct substitution, giving us that dπ¦ by dπ₯ is equal to negative one over one plus π₯ multiplied by one plus π₯. And so, we reach our solution, which is that d by dπ₯ of one over one plus π₯ is equal to negative one over one plus π₯ squared. In our final example, weβll see a different use for the definition of the derivative of a function.

Evaluate the limit as β tends to zero of π of β plus four minus π of β minus two plus π of negative two minus π of four all over β.

Here, we can see that the limit weβve been asked to evaluate looks very similar to the definition of the derivative. The definition of the derivative tells us that π prime of π₯ is equal to the limit as β tends to zero of π of π₯ plus β minus π of π₯ all over β. Letβs try to rearrange the expression within our limit to see if we can try to isolate the definition of the derivative of π at some point.

The first thing we notice is that in the numerator, we have an π of β plus four and an π of four. So, we can group these two terms together. We also have an π of β minus two and an π of negative two. So, we can also group these two terms together. Now, that weβve done this grouping in the numerator, we can split our fraction into two accordingly, giving us the limit as β tends to zero of π of β plus four minus π of four all over β plus the limit as β tends to zero of π of negative two minus π of β minus two all over β.

We can notice that this first limit is looking very close to the definition of π prime of four. If we were to write out π prime of four, we would see that it is equal to the limit as β tends to zero of π of four plus β minus π of four all over β. The only difference between this limit and the limit we found in the question is that β plus four and four plus β are the other way around. However, since the order of addition does not matter, these two things are in fact equal. And therefore, the limit as β tends to zero of π of β plus four minus π of four over β is, in fact, π prime of four.

Now, letβs look at the second limit. We can see in the definition of the derivative that we subtract π of π₯ from π of π₯ plus β. However, in our limit weβre subtracting π of β minus two from π of negative two. In order to get this in the right order, we need to multiply our fraction by negative one. Weβre able to do this if we stick a minus sign in front of our limit. We obtained the negative limit as β tends to zero of π of β minus two minus π of negative two over β.

Now, this is looking very close to the definition of the derivative of π at π₯ is equal to negative two, since π prime of negative two is equal to the limit as β tends to zero of π of negative two plus β minus π of negative two all over β. And again, we can see that this is identical to our limit except for the fact that β and negative two are the other way around. But we know that these two things are equivalent.

And so, we can say that our second limit is equal to π prime of negative two. And therefore, we have reached our solution. And that is that the limit as β tends to zero of π of β plus four minus π of β minus two plus π of negative two minus π of four all over β is equal to π prime of four minus π prime of negative two. We have now covered the definition of a derivative and a variety of examples including it. Letβs recap the key points of the video.

Key Points

The derivative of a function is defined as the limit as β tends to zero of π of π₯ plus β minus π of π₯ all over β. An alternative but equivalent definition of the derivative is the limit as π₯ one tends to π₯ zero of π of π₯ one minus π of π₯ zero over π₯ one minus π₯ zero. There are two common ways to denote derivatives β prime notation, which is π prime of π₯, and Leibnizβs notation, which is dπ¦ by dπ₯. The derivative defines a function which is equal to the slope of the tangent at each point on the curve.