BAM!
Mr. Tarrou.
So we just learned the basic rules of differentiation.
Why are they called basic rules?
Well because they are basic.
They are not always enough.
So we are going to start learning other techniques, other rules that we need, to differentiate
equations.
One of them, the one in this video, is going to be called or is the Product Rule.
When you want to find the derivative with respect to x of two functions multiplied together,
it is equal to the first function... or factor since these are multiplied together...
times the derivative of the second function plus the derivative of the first function
or factor times the second factor or function.
So we are going to have 5 examples and you will see that it can be a little bit tedious.
So we are going to this example, this first example only, twice.
I don't think you can do the other ones in multiple ways.
This first example, we can either use the Product Rule or we can use the Basic Rules
of differentiation.
Just to make sure that understand how this works we are going to use the Product Rule
first.
Then I am going to adjust this a little bit and show you that just because the problem
looks like it requires the Product Rule, you may be able to manipulate it a little bit
and just use one of the Basic Rules which may make it quicker.
So we have y equals 2x^2-5 times -3x^3+x.
So the first factor times the derivative of the second factor... so y prime, or the derivative
of y, is going to be equal to 2 x squared minus 5 times the derivative with respect
to x of -3 x cubed plus x... plus the derivative of the first factor which is here... so the
derivative with respect to x of 2x^2-5 times the second factor which is -3x^3+x.
So we have... well...
I am just going to copy this down again.
2x^2-5... the derivative with respect to x with this factor is going to be -3 times 3
which is -9... and x... we need to reduce that exponent by 1, so that is going to be
squared... and then we have plus the derivative of x with respect to x, that is going to just
be 1... plus the derivative of 2x^2-5 is going to be 4x^1... the derivative of a constant
is zero so we don't need to write that because we are just adding zero.
And I don't need to write an exponent of 1...
And times the second factor of -3x^3+x.
Now all I have to do... is well... that is all I have to do.
I have to distribute this all out and simplify until I get my answer.
We have 2x^2 times -9x^2 is -18x^4... and then positive 2.. that would be to the fourth...
Positive 2x^2... then -5 times -9 is positive 45x^2... now minus 5.
Then we are also going to distribute this 4x and get plus... 4 times -3 is...
Well let's take the whole positive 4x and multiply it through.
So positive 4 times -3 is -12x^4... and positive 4 and positive 1 is plus 4x^2.
Now I have to add like terms.
So we have -18 and -12 make -30x^4.
We have 2 and 45 makes 47.. and 47 and 4 makes 51, so 51x^2... now minus 5.
Let me make sure I did not make some small mistake here.
Good.
Alright.
So that is the derivative of y using the Product Rule.
Now, I could have just distributed this together in the first place and used the Basic Rules
of differentiation instead of this process which is a little bit tedious.
Yeah.
Ok, so we have y is equal to... let's see... 2x^2 times -3x^3 is negative 6 x to the fifth.
And then 2x^2 times x is positive 2x^3.
Then we have -5 times -3 is positive 15x^3.
And then we have -5 times positive x, which is -5x.
Ok.
Well, now we have y is equal to -6x^5... 2 and 15 makes 17, so plus 17x^3.. minus 5x.
Now I just have a basic polynomial that is very easy to find the derivative.
Y prime or the derivative of y is equal to, and this all with respect to x, is going to
be 5 times -6 is -30x^4... 3 times 17 is 51, so plus 51x^2 minus 5.
You see I got the exact same answer as I did in my first method of doing this problem.
I got it with a decent amount less work.
So if you can maybe clean up your original problem before immediately thinking I have
to apply the Product Rule, you might save yourself a little bit of time.
The next four examples though I don't believe you are going to have the possibility of doing
this.
I say that because I don't remember exactly what I wrote down for examples.
Yes.
So we are going to need to use the product rule for the remaining, I said four examples,
we have a remaining three examples.
Alright, BAM!
For our next two examples we have f(x) is equal to the square root of x^3 times the
cosine of x.
And this whatever over here.
So we are going to deal with this first.
We cannot, or do not prefer, to work with radical notation.
We are going to change this square of x to the third power to... let's see here...
Power and root, this is the square root so that is going to be x to the 3/2 power.
And now apply the Product Rule.
So f prime of x, or the derivative of f(x), is equal to the first factor times the derivative
of the second so it is going to be x^3/2 times the derivative with respect to x of cos(x)
plus the derivative of the first factor... so the derivative with respect to x of x^3/2
times the cos(x) with of course no derivative notation on it.
That is going to become x to the 3/2 power... the derivative of the cos(x) remember is -sin(x)
and I am going to let that negative float out front...
sin(x)... plus the derivative of x^3/2...
I am going to take that exponent and move it out front.
It will be 3/2 and reduce that exponent by one, 3/2 minus 1 is 1/2... times cos(x).
That is it.
The only thing, if you wanted to write this so that you are back into radical notation
you could but that is certainly acceptable and correct for the derivative of f(x).
Now moving on to this example over here.
Again we have...
Now inside this second factor we have two terms, so we are going to be doing the sum
rule and the product rule at the same time.
But it is the cosine times this second factor.
So f'(x)....
Put in a dividing line in there.
f prime of x is equal to the cosine of x times the derivative with respect to x of sin(x)+cos(x)...
now I am here so plus the derivative with respect of x of the first factor of cox(x)
times... and I am starting to run out of room... the sin(x)+cos(x).
I left myself plenty of room there.
So cosine of x.
The derivative of sin(x)+cox(x), well there are two terms in here.
We learned with the sum rule that we take the derivative of each term separately.
So the derivative of sine is cosine.
The derivative of cosine is negative sine.
So instead of plus we have minus sin(x).
Again this is plus, but I don't want to write that yet because the derivative of cosine
is again negative sine.
Let me just out of my head.
Here minus sine x times sin(x)+cos(x).
Alright, let's see here.
Distribute through the parenthesis.
We get cosine times cosine is cosine squared minus sin(x)cos(x) minus sine squared x...
and again we have minus sin(x)cos(x).
We have like terms here.
We have two sin(x)cos(x) terms so we have ... and I am going to bring this sine squared
term over...
We have cosine squared x minus sine squared x, and then -1 and -1 makes -2sin(x)cos(x).
That is the derivative of f(x).
That would perfectly fine.
I would be happy with that.
You could get super fancy because this is a double angle for cosine.
So that is the cos(2x).
And this is a double angle identity.
2sin(x)cos(x) is equal to the sin(2x), but that is perfectly acceptable and correct for
the derivative of our original function.
Let's do one more example and call it a day, shall we.
Last example.
We have f(x) is equal to -3xsin(x) plus 4tan(x).
I chose this example because in one term we have to use the Product Rule.
We basically have here -3x times the sin(x), and over here we have 4tan(x).
This is some stuff multiplied together.
If I group this up we have two factors multiplied together.
Here we have that as well, but with the tangent of x we have a constant.
We are going to use the Constant Rule for the second part of this.
You could use the Product Rule, but you are just going to write a bunch of stuff and it
is going to cancel... you are going to get the same answer with a lot more work.
So we are going to use the Constant Rule for the second term.
We cannot use it, that is not a constant...
so we are going to need to use the Produce Rule for the first one.
So f'(x) or the derivative of f(x) is equal to -3x times the derivative with respect to
x of the sine of x plus the derivative, using the Product Rule here, the derivative with
respect to x of -3x times sine x plus the derivative with respect to x of 4tan(x).
Now when we did the Constant Rule I showed you how you can take these constants of -3
here with this one and the 4 and move them out front.
I am going to save myself the writing and not do that again.
We have -3x times the derivative of the sine of x is cos(x)... plus the derivative of -3x
is simply equal to... it is -3 x to the first.
So bring that down that 1, multiply it to 3, and get -3.
The exponent goes down to 0, so the x becomes 1.
x to the zero power is one.
So the derivative with respect to x of -3x is -3... times the sine of x...
This 4 is going to come out and we get plus 4... and the derivative with respect to x
of tan(x) is equal to secant squared x.
And that is pretty much it.
I guess I could...
We don't...
Why write plus negative?
So we can take that and just write minus, a big ol minus sign, 3sin(x).
That is it.
That is the answer.
With your Trig functions you might get what you feel is a correct final answer, look in
the back of the book, and not see your answer.
These parenthesis are not doing anything because there is just one term inside.
So you might get what you think is a good final answer and it is.
It is correct.
This is f prime of x, or the derivative of f(x).
But then you might go in the back of the back of the book and ask, "Where did the secant
go?"
You know secant squared x is the same of 1 over cosine squared x.
So I could make this 4/cos^2(x), and then find common denominators and all of that stuff.
So you might do your derivative problems correctly that have trig functions in them, but then
not see the exact form of your answer in the back of the book.
It just makes it difficult to make sure that you have correct answer.
So this is good.
If you go in the back of the book, or whatever problem you are working on, and that is not
what is back there, maybe just take what you have as an answer and try to manipulate to
appear the same as what is in the back of the book.
It does not necessarily mean that you have done it wrong.
I am Mr. Tarrou.
BAM!
Go do your homework:D