Last time, we discussed the concept of equilibrium.

Now, equilibrium is very important concept in economics.

One way to understand equilibria is to

understand what happens if condition in the markets are changing.

Now because the condition in markets are shifting all the time,

it's very important for us to understand what the

effect the change has on the equilibria conditions.

In order to do that,

I will need to introduce you with the concept of derivative.

Now, you have already studied derivatives in one of your lectures.

What I'm going to do right now is do a little bit of overview of what derivatives are and

how they can be seen in the graph in order for you to grasp the principles much better.

So let's take a linear function as an example, any linear function.

Again, we have the same axis',

axis y on the vertical line and axis x on the horizontal line.

And let's take just a simple linear function where y equal x.

Now, what is the rate of change?

What does it mean to have a rate of change?

Now, in principle, if we take any two points,

let's take say two and three,

and find the corresponding values of y.

And this case is quite straightforward,

we're just plugging the value of two and the value of three

here and we get that when x is equal two,

the value of y will correspond to that, it's also two.

And when x is equal three,

the value that correspond to that is also three.

Now, the rate of change will say and you or will actually answer the following question.

What will happen to your function y whenever your x's are changing?

In this example, by one unit only.

So here is quite straightforward and we can actually

see that because we pick a quite straightforward linear function,

the rate of change when x is changing only by one unit,

the rate of change in y will also be one unit.

Delta here and delta here,

it's just saying the rate of change or the change,

actually, not the rate of change.

Now the derivative is actually asking or answering the following question.

What will be the change in y as a result of change in x?

In this example, like I said,

we have a change in x,

which was only one unit and y,

the result was also one unit.

So the rate of change was actually one.

But in this very linear function,

it's quite straightforward to see or quite easy

to see that any one unit that will take so,

for instance, seven and eight.

The rate of change of y will be exactly the same as the rate of change here.

Things started to be a bit more complex,

when we take a slightly more peculiar function than this.

Let's pick up a more elaborated function.

Let's pick up y equals 3x.

For the same rate of change of one unit in x,

it will be much more steeper function.

Although it's a stepper function,

if you plug back the original values, what will we see?

So let's plug back original to do the original value of two and three into this function.

So for y, when we have x is equal two,

we have y equals three times two, which is six.

And whenever yx equals three,

what we have is,

three times three, which is nine.

Now the rate of change is the result in

y will be nine minus six, which is three.

If we take these values of seven and eight and plug them back,

we'll get exactly the same outcome that the rate of change will be also three.

I invite you to do this at home to see whether

you understand the concept of rate of change.

So the answer to the following question,

what will be the rate of change in the linear function like that of y equals 3x,

we'll say that y is changing by three when we change in x by one unit,

and the rate of change will be three.

Any other linear functions can take,

y equals 7x, y equals 8x,

and so on and so forth.

You understand now that the rate of change is represented by this value alone.

Now, if you're familiar a little bit with derivatives

you can already see why this is the case because if you

remember with derivatives you said derivatives that can be represented

by symbol y_prime is equal to three in this case,

seven in this case,

and eight in this case.

We will see in a minute why this is indeed the way that we find the derivative.

But things are starting to be much more complex when we don't take linear function but

much more advanced functions like a parabola or quadratic function.

Let's start to understand that slightly more

in order to understand a derivative a little bit more.

So if we get rid of that and plot the new function,

let's say quadratic function so like before we plug the quadratic function by,

first of all, putting this in the axis.

Let's take a quadratic function,

which is y equals x_squared.

Now, if we take the same one and two

here and let's say three and four here,

the rate of change of x remains the same.

The change in x, I would say, remains the same.

But, what happens to y?

And this is a very interesting question.

Let's try and find this out.

So, first of all,

in order to find the change in y with

corresponding values of one and two and three and four,

we have to plug them into the original function.

So when the y axis equal one,

we just plug one_squared, we get one.

When x is equal two,

we square the two and we get four.

Now, as a result of that,

the change in y is four minus one, which is three.

If we take the values of three and four,

you can imagine that the distance between

the value of y here and here will be slightly different.

Let's find out how much it is.

Y when x is equal three,

we square this three,

we get the nine,

yx equals four, we get four_squared and we get 16.

Now, the rate of change in y here is slightly different.

So 16 minus nine is equal to seven.

So, unlike linear functions,

which had the same change in y regardless where you take the change of x of one unit,

in function like this,

the change in y will be different and will depend on the value of x's here.

Now, if we can think about this as approximation to linear function so, for instance,

if we take those four points and try to plot a linear function with them,

we see that with these two points,

the linear function looks something like that.

However, with these points,

the linear function would be much,

much steeper as you can imagine.

Let's put this again.

So you can clearly

see now that the further away you go from the original point zero here,

the change in y as a result of a change in one unit of x, will increase.