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G'day.
Welcome to this fourth and last video in my series
on how to find the derivatives of compound functions.
In my previous three videos, I've dealt with the Chain Rule,
the Product Rule and the Quotient Rule and, in this shorter video,
I propose to show how you can use those three structures
to find the derivatives of more complicated functions.
Now, I'm not going to go through the theory of the six basic functions any more.
I assume you've looked at those videos and you know what I'm talking about.
All I propose to do is to [is to solve]
find the derivatives of two compound functions for you
and to show you how the structure helps you find those derivatives
with considerable ease and speed.
So, without wasting time, let's see how the structure helps.
I'm going to create a compound function here,
with a product on the top ... let's make it e^(6x)ln(x+1) ...
and something fairly complicated on the bottom, like x^3cos^5(7x) ...
and I think you'll agree that's a good deal more complex
than most questions, or perhaps any question,
you have in your school textbook!
Perhaps, if you're at university, you might find this useful as well.
But, how do we find the derivative?
Actually, I'm going to retain the blue pen and set the structure up.
Remember, since the overall picture here is a quotient, or a fraction,
where we have one function over another, so I'm going to set up a structure
with a very long vinculum and square the bottom.
Now, squaring x^3 is easy,
and squaring a cosine to the fifth power will give us cos^10(7x),
so we've squared the denominator!
We know it's going to involve a minus sign, because it's a Quotient Rule.
And, we're going to have the derivative of the top
multiplied by the bottom, so I'm going to write the x^3cos^5(7x) ...
and, in the second part of the expression,
we're going to leave the top alone
and multiply by the derivative of the bottom.
So, what I've done is ... I've already drawn out the structure
that takes care of the quotient part of the derivative.
Now, when we find the derivative of the numerator, to put it here,
we have a product of two functions ... it's a Product Rule.
So, I know that I'm going to have two expressions added together.
In fact, I know more than that!
I know that I'm going to find the derivative of e^(6x)
and multiply it by ln(x+1) ...
and then I'm going to leave the e^(6x) alone
and write down the [derivative of the] ln(x+1) there.
Similarly, when I come to find the derivative of this denominator,
it also is a product ... of x^3 times cos^5(7x).
Consequently, its derivative is going to appear
as the sum of two terms, or two expressions, in here.
And, I'm going to have the derivative of x^3 times cos^5(7x)
... plus x^3 times the derivative of cos^5(7x).
What I've done now is I've put the structure for the Product Rules here.
So you can see ... first of all I dealt with the Quotient Rule,
then the Product Rule ...
and, now, all that I'm left with are small Chain Rules ... small chains.
Here I must do the derivative of e^(6x) which is itself ...
so the derivative of the exponential is itself ...
times the derivative of the function inside,
which is the derivative of 6x ... is 6 ... times whatever's left.
Then I leave this alone and I find the derivative of ln(x+1).
Well, the derivative of ln(x) is one over whatever's inside
times the derivative of whatever's inside.
When I come to find this, I find the derivative of x^3 is 3x^2 ...
times cos^5(7x) ... plus ... the x^3 I leave alone ...
and then I find the derivative of cos^5(7x) ...
which is actually a little bit longer than this, so I should leave ...
should have that bracket a little bit further over ...
I deal with the fifth power first, so it's going to be 5cos^4(7x) ...
times (I might have to work above here) ... times ...
the derivative of cosine is –sin(7x) ... and then multiply by
the derivative of 7x which is 7.
And there it is in one line!
We've completed the derivative of that compound function.
Now, of course, you have ... this is quite an horrific example ...
and you would have a substantial amount of sorting out to do, algebraically.
But, in fact, the derivative has been found.
And I want you to learn to use this structure so that,
with simpler examples,
you can find derivatives very, very quickly indeed.
Let's do one more.
(Sorry)
Let's do another quotient ... ln(x)...
I'm trying to give a variety of functions here ...
e^tan(7x), for example ...
and, on the bottom, we might put x^3sin^2(5x).
So, how do we find the derivative of this compound function?
Again, a very long bar ... very long vinculum ...
we square the bottom ...
I just realised that's not very different from what I had
so it just means doubling those indices.
We draw the minus sign in, and we're going to have
the derivative of this times this ... so I'm going to write my x^3sin^2(5x) ...
and then we're going to have the numerator, ln(x)e^tan(7x),
times the derivative of what's on the bottom.
So, our structure has been put in place for the Quotient Rule.
Now, let's deal with this product on top.
We have a product of two functions, so I'm going to put brackets ...
By the way, the square ones are genuinely brackets.
I'm a traditionalist. I learned the original names of things.
The bent ones are parentheses and the curly ones are called braces.
People who never learned the names for those particular grouping symbols
call them curly brackets and square brackets
and curvy brackets, or something,
but they were originally parentheses, brackets and braces,
and you'll just have to put up with my using those terms.
Now, since we have ... we're finding the derivative of a product ...
we know we're going to have two expressions added ...
we're going to find the derivative of ln(x) times e^tan(7x),
and then we're going to leave the logarithm alone
and find the derivative of that.
Over here, we're going to find another product,
because I have a product on the bottom.
We'll find the derivative of x^3 times the sin^2(5x) ...
and then we'll leave the x^3 alone and find the derivative of sin^2(5x).
Now, I really don't have room to fit these in,
so I'm going to stack them a little bit with my red ink.
What's the derivative of ln(x)? Well, it's quite simple!
So, we find the derivative of ln(x) is 1/x ... times what remains ...
leave the ln(x) ... derivative of this ...
well, the derivative of e to anything is itself ...
then we multiply by (I'm going to have to squeeze this in above) ...
we multiply by the derivative of the tangent which is sec^2(7x) ...
and then we multiply by the derivative of the function inside that,
which is 7.
I hope you can see that this structure gave us the Quotient Rule,
this structure gave us the Product Rule (here and here) ...
and now we're building in the Chain Rules.
[The] derivative of x^3 is 3x^2, and then we leave the sin^2(5x) alone ...
leave the x^3 here, and the derivative of sin^2(5x) is 2sin(5x) ...
that deals with the power of 2 ... times the derivative of sine is cos(5x)
... times the derivative of the function 5x inside.
Now, again, you have a great deal of sorting out to do there ...
expanding grouping symbols and, in some cases, gathering like terms
and such like ... a rather messy, nasty bit of algebra there.
What I've demonstrated in these two examples, here,
is how to find the derivatives of quite complex compound functions,
and I hope that's been of help to you.
Now, you're not going to find a great number of these
in your school textbook.
If you can get access to university level textbooks, you will,
and you may be able to find some sources of them on the Internet.
You might find that your teachers have access to sets of them.
Wherever you source your examples from, I encourage you to practise
finding the derivatives of compound functions like this
because, if you can do this, you'll certainly be able
to find the derivatives of the most complicated functions
they'll give you in your external examinations at school.
Thank you very much for watching.
I encourage you to subscribe to my channel
to find out about future videos.
And please leave a comment and 'like' this video as well.
Thank you.