G'day.
Can you find the derivatives of products of functions
in just 20 or 30 seconds, or even less?
If you can't, then you need to watch this video.
In it, I'm going to first of all demonstrate how it's done,
then I'll show you the basic theory behind it,
and how to learn it, and how to put it into practice.
Here, I have behind me, six examples ...
six equations where we have a product of functions in every case.
They're from the New South Wales Higher School Certificate in Australia
which is the basic level advanced course for school leavers.
And these are questions towards the beginning of the paper,
so they're easier questions, worth two marks each, and therefore,
the time allocation is about three minutes per question.
We're going to, first of all, solve them without much discussion
and I'm going to demonstrate
the kind of speed that you should be able to achieve
in your Higher School Certificate examination
or whatever examination you're sitting for around the world.
Let's have a look.
The derivative of this product is the derivative of that ... (x-1)
... and the derivative of that, of course, is 1 ...
times the ln(x) plus ... leave this alone ...
multiplied by the derivative of ln(x).
1 times ln(x) of course is ln(x) and (x-1) times 1 is (x-1) ... over x.
So, we have the derivative stage and the simplification stage.
Now, you can time this.
I'm certainly not going to take three minutes per question.
And, with practice, you should be able
to do the first step in about 10 seconds, and then use another 10 seconds or so
to simplify the expression (or the equation).
The derivative here will be 2x times tan(x) plus ...
leave the x^2 alone and multiply by the derivative of tan(x) which is sec^2(x)
... and, in fact, you really can't simplify that much
beyond taking x out as a common factor.
The derivative of this one will be 2x times ln(x) plus x^2 times 1/x.
Simplification ... 2xln(x) ...
and one of these 'x's will divide out, leaving you just the simple 'x.'
By the way, this was the 2012 HSC question ... 2010 and 2008.
Here's the 2011 question ... plus x^2e^x. That's worth two marks.
It's supposed to take up to three minutes. I think we took about 10 seconds.
This one here ... (sorry) the derivative of x is 1 ... times sin(x)
... plus x times the derivative of sin(x) is cos(x).
And the only tidy up you'd need is ... we'd normally wouldn't write
1 times sin(x) ... that was worth 2 marks.
This one here ... the derivative of x is 1,
so it's 1 times tan(x) ...
plus x times the derivative of tan(x) which is sec^2(x)
... which is tan(x) + xsec^2(x).
Now, I suspect that took us about 2 minutes to complete the page.
If I didn't have to talk, we'd probably reduce that
to not much more than a minute for all six questions!
What I propose to do now is to explain the theory
behind this very quickly,
then show you a simple structure that you should memorise
that will help you be able to solve all these equations ...
or at least find the derivatives for all these functions,
and then I'll quickly demonstrate a few examples afterwards
... large examples as well ...
so that you understand the principle very thoroughly.
So, what are the foundations of the Product Rule?
For school students in my state in Australia,
there are six basic functions.
They must be able to find the derivative of x to some power
... which is the foundation of all polynomials.
They must be able to find the derivative of an exponential ...
the derivative of a logarithmic function, the derivative of a sine function,
a cosine function, and a tangent function.
If you can write a summary of all these six functions and their derivatives,
then you've already established the foundation for using the Product Rule.
The derivative of x^n is nx^(n-1),
that is, the index comes down the front and then it's reduced by one.
The derivative of the exponential is itself.
The derivative of the logarithmic function is 1/x.
The derivative of the sine function is cos(x).
The derivative of the cosine is –sin(x).
And the derivative of the tangent is sec^2(x).
Now, how does that work with the Product Rule?
Well, let's make a product of some of these functions.
Let's write y = (let's say) e^x times sin(x) times ln(x) times tan(x).
Now, each of these individual functions is not complicated.
I'll deal with that in another video but we do, in fact,
have a product here of four functions. Let's look at the derivative.
When I was at school I was taught
how to find the derivative of the product of two functions.
We called them u and v and then the derivative was u'v + v'u.
I find that students get confused with that.
I want you to see the pattern simply as a structure on a page
... and here it is.
The derivative of this is the derivative of e^x
times the remaining functions.
If you like, this is our u'v ... plus ... then we move to the next function
... we leave the e^x alone ... we leave the ln(x) and the tan(x) alone
... but we find the derivative of sin(x) which is, of course, cos(x).
Then we move to the next function, which is the logarithm.
We've done the derivative of the exponential,
the derivative of the sine function
... now we work on the derivative of the logarithmic function
and we leave the other three alone, so I'm going to leave them in blue.
The derivative of the logarithmic function is 1/x.
And then the final step is to move across to the tangent function,
seeing that we have already done
the exponential, the sine and the logarithmic functions.
We'll find the derivative of tan(x) and leave the other three alone.
And, the derivative of the tangent, of course,
from our little summary at the top, is sec^2(x).
Believe it, or not, that is the derivative of that compound function.
The Product Rule, as you see, when you get past the product of two functions,
looks a bit like a grid, or a matrix, with a diagonal composed of
the derivatives of each of the functions.
So, if you had six functions, you would have six separate expressions
all added together and, in each expression,
everything would remain the same except that one of those six functions
would be replaced by its derivative.
I'm going to do that one more time and then we're going to look at
how to apply that to smaller and simpler examples.
So, let's try a different function.
We didn't use a power of x before, so let's use x^5.
I don't think we used the cosine function before ... cos(x)
... e^x ... sin(x) ... let's make it tan(x) as well.
Here we have five functions, so I think you will appreciate
that we're going to have five steps underneath.
The derivative is going to have five lines
which I'll try to write very quickly.
In fact, let's write the blue material first.
We're going to, first of all, work out the derivative of x^5,
so we're going to leave the cos(x)e^xsin(x)tan(x) in place.
[That's awful ... there we go ... that's not much better ...]
Then we're going to leave the x^5 in place
and work out the derivative of cos(x) and leave the e^xsin(x)tan(x).
Then we're going to leave the x^5cos(x).
We're going to work out the derivative of e^x
and leave the sin(x)tan(x).
And then we're going to work out the x^5cos(x)e^x ...
and the derivative of sin(x) will go here ...
and then we'll have tan(x).
And, finally, x^5cos(x)e^xsin(x) and the derivative of tan(x).
So, we've now got our one, two, three, four, five lines
all prepared for the derivative.
The derivative of x^5 is 5x^4
(times whatever's left, which we've already written),
the derivative of cos(x) is –sin(x) (times the remaining four functions).
The derivative of e^x is e^x
(and we've left the remaining four functions to be multiplied by it).
The derivative of sin(x) is cos(x), and the derivative of tan(x) is sec^2(x).
And that is the derivative of that function.
That's the Product Rule in action, without using substitution ...
without using u'vs and v'us. That's it!
So, I'm going to clear the board now and, in the remaining minute or two of the video,
we're going to look at perhaps four examples and see how we use this particular process
to find their derivatives.
Well, I've decided to find the derivatives of five functions instead of four
and here they are.
When we've worked through these,
I'd encourage you to go back to the beginning of the video and
... stop when you wish ...
but look through the first six examples that I gave from HSC examinations
in my state in Australia and see this process in action
on those questions.
But here we go ... the derivative is
the derivative of the product of these functions.
So we do the derivative of this function first, which is 5x^4,
leave the e^x alone,
then we leave the x^5 alone and find the derivative of e^x.
Now, of course, you could if you wish, take out the common factor of x^4e^x,
and this would leave (5+x) ... or (x+5).
This one? The derivative.
We have a sine function times a logarithmic function.
The derivative of sin(x) is cos(x), leave the ln(x) alone ...
I'll write underneath ... the next expression,
we leave the sine function alone
and we find the derivative of the logarithm which is 1/x.
I'll just leave that as it is.
This derivative. Cosine times an exponential.
Derivative of cos(x) is –sin(x), leave the exponential alone,
and then we add on ...
the cos(x) times the derivative of the exponential.
So, the first time around we do [find] the derivative of the cosine,
the second time around we do [find] the derivative of the exponential function.
Here we go. The derivative of this product
will be the derivative of x^4 times the remaining tangent function
plus ... leave the x^4 alone ... and the derivative of tan(x) is sec^2(x).
And, finally, I decided to add this one which is a product of four functions,
so we're going to have four lines.
The first one, the derivative of sin(x) is cos(x)
... leave the rest alone.
And, you can see that, with time, you can do this very, very quickly!
And now, I find it helpful to point with my finger.
So, we've dealt with this. The next one, we're dealing with this.
So, we leave the sin(x) alone ... derivative of e^x is e^x ... ln(x)x^7.
... moving across ... The next one is sin(x)e^x ...
the derivative of ln(x) is 1/x ... times x^7.
And [for] the final one, we leave the first three functions alone
... and find the derivative of x^7 which is 7x^6.
Now, of course, when one tidies this up, this little product here
could be simplified ... the x would divide into x^7 to create x^6.
There are some common factors you could take out
but I'll deal with that in another video.
But here, I've just concentrated on
showing you how to find derivatives using the Product Rule,
and I hope that's made sense to you.
Let me encourage you to practise, using examples from your textbook,
or from the Internet, or wherever you can obtain them
... past exam papers.
Pester your teachers until they give you pages and pages of them.
And practise them until you can (solve these) find these derivatives
in at least 10 or 20 seconds and, hopefully, even less.
So, if you can find these derivatives at the rate of 3 or 4 per minute
then you can appreciate that, every few weeks or every few months
you can pull out an exercise of them and do 20 or 30 or 40 examples
in a matter of 10 minutes ... and that is very, very good practice.
So, I hope that's been of use.
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Thank you for watching.