Differentiation Rules

Hello. I'm Samuel Siksik,

professor of math at the University of Warwick.

This is the second in a series of courses on

differentiation.

We're going to start out by looking at some properties of derivatives.

So what you see in this table is functions are made up of other functions,

so I have y which is equal to K times f of X l k is a constant f of

X is the function or I have wise equal to f of X plus g of X and so on.

And in the second column,

I have

the derivative of Why so if I differentiate k times f of X k is a constant,

I get k times the derivative

of F. So I get k times f dashed of X.

If I did differentiate the some f of X pastilles of X, I get f dashed X plus D dashed of X.

If I differentiate the difference f of X minus D of X, I get

f dashed of X

minus g dashed of X. So the first three

properties or rules are

very intuitive.

The next two of those when I differentiate product

or cautioned there a little bit less intuitive,

and you have to be more careful when you apply them.

So the first one is called the product I'm differentiating.

Why is equal to F of x Times D of X

and the derivative is

f dashed of x times g f x

plus f of x times G dashed of X.

This is called the product tool,

and then there's a caution tool.

When I'm differentiating, y is equal to F of X, divided by G f X

and I get the derivative is equal to the caution of

in the numerator. I've got g f x f dashed of X minus f of X g dashed of X,

and in the denominator, I have G f X, all squared.

So these are two of those that you should be familiar with,

that you should be at home with and apply them efforts.

You will need them again and again.

So to help us be familiar with the rules,

let's look at an example to the example says determine the points on the graph.

Y is equal to X squared

minus line of X,

where the gradient is one

and to start out.

I have to remember that gradient is the same thing as the derivative.

So I want to differentiate. Why is equal to X squared minus Leonard X.

I'm going to need two things.

One of them is that all for differentiating a difference. So I'm going to take

I'm going to look at this level, which is y is equal to f of X minus G of X.

It tells me that d y by D. X is equal to f dashed of X minus G dash

x. I'm going to take my f of X to be X squared

and my g of x to be lin of x.

Okay,

what's the derivative of X squared?

Remember the derivative X to the N is equal to end times X to the end minus one.

So the derivative of X squared is two X

What's the derivative of Lin X?

So I'm going to apply the blow Where I take K is equal to one.

So the derivative of Lin X is one over x.

Okay,

Now I apply the difference rule So the derivative of X squared minus lin X is D Y by D.

X is equal to two x minus one over X.

Okay,

so it's worth pausing the video and looking at this again and making

sure that you understand where we got the formula for the derivative from.

Okay, now that we have the formula for the derivative, remember,

what we want is to know where the gradient is one.

So that means where is the derivative equal to one.

So I equate my formula for the derivative, which is two x minus one over X.

I equated to one

and I multiply by X,

I rearrange and I end up with a quadratic equation.

Two X squared minus X minus one is equal to zero. And here's another.

You know, maybe you want to pause the video now and try to solve this on your own.

Okay,

But if you factor that you were going to get to

expose one times X minus one is equal to zero.

So that tells me that X is equal to minus a half or X is equal to one.

So that's the X coordinate

I need also the y coordinate because it said determined, the question said,

determine the points on the curve Y is equal to X squared minus Lin X,

where the gradient

So I want to determine I want to substitute X is equal to minus a half in the formula Y

is equal to X squared minus Lin X and I

want to substitute X equals one the same formula.

Now I'm not allowed to substitute minus a half

in Lynn because Lynn only takes positive values.

So I exclude X is equal to a half. I've got X is equal to one

and this gives me that y is equal to one squared minus Lynn of one

lane of one is zero. So I get Y is equal to one.

So the only point where the gradient is one

is 11.

Now, as an exercise, here is a series of functions.

Um,

you're asked to differentiate them and what you're going to need is the two tables.

One of them is the table of the derivatives of standard functions which you see

on the left and also the rules for differentiation which you see on the light.

In this video we saw some rules for differentiation,

and we've also seen an example of how to apply them.

And I've left you with an exercise to do,

which will help you get to grips with these rules.

In the next video, we're going to look again at

a couple more examples just to solidify our appreciation of these laws.