BAM!
Mr. Tarrou.
So we are looking at the Chain Rule.
I did a video introducing the basic concept and rule and the General Power Rule and then
said I would do a couple more advanced examples involving algebra.
Then we are going to move into a whole other video where, or two videos maybe, where we
are going to do the Chain Rule involving trigonometry functions.
We have y equals x+2 over x^2-5 raised to the third power.
So we got a big ol chunk of something here raised to the third power.
We are going to use the General Power Rule and say that y prime is equal to 3 times x+2
over x^2-5 and now that is going to be raised to the second power because we of course bring
that power down by one.
But this is the Chain Rule.
So we just did this first part.
Now we have to find or multiply by the derivative of the inside function, or the derivative
of U.
We are going to... well... let's see.
Let's show all of the work.
We have the derivative with respect to x of x+2 over x^2-5.
Now to finish up this Chain Rule I have to find the derivative of a quotient.
In case you forgot, we have the derivative of a quotient is equal to the denominator
times the derivative of the numerator minus the derivative of the denominator times the
numerator over the denominator squared.
So we have 3 times x+2 over...
now I am going to go ahead and also apply this power of two to the numerator and denominator
of the fraction.
I am not actually going to distribute it out.
I am just going to write x+2 squared over x^2-5 squared.
Let's go ahead and lose those parenthesis and just put the 3 on top.
This is 3 over 1.
Ok.
So here we have the derivative of x+2 over x^2-5 is as we can see in the quotient rule..
we are going to take the denominator of x^-5 and multiply it by the derivative of the numerator.
So when we take the derivative of x we just simply get 1.
This is a constant so the derivative of x+2 is simply 1.
Now minus, says over there g prime of x... the derivative of the denominator.
We are going to have x^2-5, this is the going to be 2x when we find the derivative here.
When you of course take the derivative of a constant it is zero, so the derivative is
just going to be 2x... now times the numerator so we have x+2.
This is all over the denominator squared.
Ok.
Well we can see that the through this multiplication we have the same denominator in both of these
factors.
We are going to just write that to the... let's see.
Any time you multiply like bases and they have ... or any time you multiply like bases
you take their exponents and you add them.
So we are going to have 3 times x+2 squared all over x^2-5 raised to the fourth power
from doing 2 plus 2 is 4.
And let's take this entire negative 2x and distribute it through the parenthesis.
I am trying to pick up red chalk, but this camera does not pick up red chalk very good
so let's get rid of that.
We have parenthesis x^2-5, let's see here... minus...
I forgot my parenthesis there...
minus -2x times x is going to be negative 2 x squared.
Negative 2x times positive 2 again is going to be -4x.
I don't really need these parenthesis around the x^2-5 because there is no leading coefficient,
there is no exponent, so this is just sitting here.
Let's take that out.
And we are going to combine like terms that are in this parenthesis.
We have 3 times x+2 squared.
We have x^2 and negative 2x^2, so it is going to be negative x squared.
We have no other single degree terms, so we are going to just have minus 4x.
Then minus 5.
All over x^2-5 to the fourth power.
I unfortunately like my students are prone to making the occasional small mistake, especially
when I am talking, thinking, and writing all at the same time.
So just to make sure I am right, here we have 3 times x+2 squared, and then in parenthesis
-x^2-4x-5 all over x^2-5 to the fourth power.
That means we are done with the first example.
One more.
BAM!
Last example.
We have f(x) is equal to 2x^2+1 raised to the fourth plus x all within its own power
of 2.
This is just going to be the chain rule done over and over and over again as we work our
way... normally with parenthesis you are starting in and working your way out.
Well, we are going to have to the reverse.
We are starting with derivatives, it is kind of like we are starting with this outside
power and we are going to work our way in until we take care of all these powers that
are inside this big ol something squared.
Let's start with that General Power Rule.
F prime of x is going to be equal to... now this is all my something, ok.... this is all
something raised to the second power.
I am going to take this power of 2, drop it out front getting 2 parenthesis...
another parenthesis 2x^2+1 to the fourth power plus x...
I am just copying what is inside this parenthesis, what had the power of 2.
I am going to take that power of two and reduce it by one.
We don't need to write that.
Now the Chain Rule.
We have just done...
you know...
We took the General Power Rule, took the 2 out front, took that power of 2 reduced it
by one... but now we have to talk about the derivative of the inside function.
That is going to be the derivative with respect to x of what is inside... 2x^2+1 raised to
the fourth plus x.
Ok.
So now this is done.
I am going to leave that along.
Now I am going to start working here.
We have 2 parenthesis 2x^2+1 to the fourth plus x.
The derivative of this.
Well we have the derivative of two terms.
The last term is simple.
The derivative of x is 1.
So that is going to be one.
But we have another big something here.
This is going to be like again our U, where we did have all of this as the inside function
and now we are doing Chain Rule again with another inside function of 2x^2-1.
So the General Power Rule again.
We have 4 times...
well ok... 4 parenthesis 2x^2+1... that General Power Rule with 4, brought it down out front,
now this get reduced by one and it is 3... but again this is something raised to the
fourth power so Chain Rule again...
This is going to be.. well I am not going to work it out.
I am showing all of the steps.
We are going to find the derivative of this inside function now.
That is going to be 2x^2+1.
Now all of this is just the derivative of 2x^2+1 to the fourth.
We have not taken care of the second term yet which is x.
The derivative with respect to x of x is 1.
Ok.
Oh my gosh.
I still have derivative notation.
[laughing] So we are going to have 2 times 2x^2+1 to the fourth plus x times 4(2x^2+1)^3...
the derivative...
times... the derivative of 2x^2+1.
Well that is a constant so it is going go be zero.
But the derivative of 2x^2 is 4x.
And then again, this is going to be the derivative of a constant which is 0.
So this 4x is the derivative of that, and then we have plus one.
Ok, so let's clean this up a little bit.
We have 4x times 4 is 16x.
I can put those two together.
We have 2((2x+1)^4 + x)... 4 and 4 is 16x... (2x^2+1)^3 + 1.
Ok, now your teacher may be totally happy with that.
But... um... it won't take a tremendous amount of work to distribute this together.
This looks like a big ugly mess, but what we have here is something plus something...
ok... and these parenthesis lock all of that together.
Another big something.
All of this is being multiplied together or has parenthesis grouping it together, so these
two groupings... these two factors are just a binomial times a binomial.
And I might not bother putting these together except that we have 2x^2+1 and 2x^2+1.
Like what has the powers...
you know... the bases are the same.
So it won't be too nasty to multiply this together.
We have 2 times... now there is a one out front here... let's just fill that in for
you.
Ok.
So we have 1 times 16x is 16x.
Then of course...
You guys know what I am doing here.
I am distributing these together, ok.
We have 1 times 16x is 16x.
Then 2x^2+1 to the fourth and 2x^2+1 to the 3, we are going to add those exponents with
the like bases.
We have 2x^2+1 to the seventh power.
Now we are going to have this times 1, so that is just going to be plus 2x^2+1 to the
fourth.
And we have the second term now x times this first term.
That is going to be x times x is x squared.
That is all we can put together.
We have plus 16x^2 times 2x^2+1 to the third.
Finally we have x times 1 which is x.
Then we can take, if we really just want to get this completely expanded as much as I
am willing to do.
I am certainly not going to raise a binomial to the seventh power.
We have a final answer of 32x times 2x^2+1 to the seventh plus 2x^2+1 to the fourth plus
16x^2(2x^2+1)^3 plus x.
Let me just make sure that I have not made any small mistakes here.
Looks like I made one somewhere.
Oh! [laughing] I would say so.
2 times 1 is 2.
I forgot to keep multiplying my 2 through.
2 times 16 is 32.
And 2 times 1x is 2x.
So I am Mr. Tarrou.
That is your last.. you know.. we will call it a fairly lengthly chain rule problem that
involves algebra.
Now let's take a look at when they involve trig functions.
I am Mr. Tarrou.
BAM!
Go do your homework :D