BAM!!! Mr. Tarrou. So we have found derivatives using the basic rules. We talked about the
Quotient Rule. We talked about the Product Rule. Yet there is still some problems that
we are not capable of finding the derivative of...some functions. We are going to bring
in another rule called the chain rule. If y equals f(u) is a differentiable function
of u, and u=g(x) is a differentiable function of x, then y equals f of g of x....We have
a function nested inside of another function. I will do an example to show you how does
this differ or what does that look like and compare with our old questions we have already
taken a look at....is a differentiable function of x. So the derivative with respect to x
of f of g of x, this is going back to our old concepts of composite functions, is equal
to f prime of g(x)...we found the derivative of the outside function and left the inside
function alone...and then multiply by the derivative of g(x). We actually take that
inside function out and find it's derivative. Or we let u equal g(x). We can do a substitution
like this to make the question...it is not really necessary unless rewriting the function
to make it look simpler helps you understand how to work through the problem... So you
can do a temporarily substitution where you say, ok let's let g(x)...maybe it is a really
complicated function inside of another one...let's let g(x) just simply equal u. Well then we
have y prime, the derivative of y, is equal to f prime of u times the derivative of u.
So you will have to go back when you are done and plug back in that original more complicated
expression that you substituted out. But, maybe this process will help you finish the
problem more accurately and easily. I probably will be doing most of my examples in this
format though. What we are doing is we are finding the derivative of y with respect to
x and we are walking through that middle function. We are finding the derivative of y with respect
to u. That is that inside function and then we are multiplying the derivative of u, the
derivative of that inside function with respect to x. We are going through this chain until
we get that derivative of y, the derivative of the original function with respect to x.
So, what are these examples going to look like? Let's find out!!! BAM! In our first
two examples we have here, f(x) is equal to x squared and f(x) is equal to parenthesis
3 x squared minus 4x squared. Well, the first one is very simple. That is just the power
rule. f prime of x is going to be equal to 2x. The second example here I have something
going on inside this squared function. So this is a function inside another function.
We have not had that before. So if I were to solve this, not solve it. If I were to
find the derivative of this without the chain rule, then I would have two options. One,
and this the first one I am going to do, is this binomial is being squared. I can write
3x^2 minus 4x times 3x^2 minus 4x. Now if I was not about to go over the chain rule
with you I would distribute this making a polynomial then find the derivative. But,
I want to match the format of the answer I am going to get when I use the chain rule
so I am going to leave this like this and use the Product Rule to find this derivative.
That is going to be f prime of x is equal to 3x^2 minus 4x times the derivative with
respect to x of the second factor so that is going to be 3x^2-4x. I am just showing
all of the steps here. Plus now I am going to take the derivative of the first factor.
The derivative with respect to x of 3x^2-4x times 3x^2-4x. Now this derivative we have
here is 3x^2-4x times the derivative of 3x^2 minus 4x is going to be 6x minus 4. All this
plus the derivative of 3x^2-4x is again we are using power rule and constant multiple
rule, this is going to be 6x again minus 4 times 3x^2-4x. Well now I have got two terms
that both have the same factors in them. They are in a different order but multiplication
is commutative so we have something times something else plus those same two somethings
still being multiplied together. I could say xy plus yx, that would be 2xy and indeed that
is what we have here. This comes out to be 2 times 3x^2-4x times 6x-4. Now that I have
found the derivative using a rule that we already know, let's apply the chain rule and
see what happens. We are going to use this example again f(x) is equal to (3x^2-4x)^2.
This portion...this is my u. Ok, that is the function that inside the other function, this
power of two function. This is my u, or this is my g(x). This outside function, this something
squared, that is my f(x). Let's write that as a y and make sure that you don't get confused
with my notation here. Ok, so y prime is going to be equal to... I am going to use the Power
Rule because this is something squared. So just like here something squared, I am going
to take the power of two and drop it down and move it out front and then reduce the
exponent by one. We are going to have 2 times 3x^2 minus 4x and that is raised to the first
power which does not do anything but I am writing it because I am teaching. That is
the f prime of g(x), now I have got to take the inside function out and find the derivative
of that. Again, I am going to write this out with all of the tiny little steps because
you are just learning it for the first time. I am assuming that is why you are watching
this video. So we are going to take that inside function out and find the derivative with
respect to x of 3x^2-4x. Ok, so now we have got 2 times 3x^2 minus 4x and we are going
to find the derivative of what was the inside function...the derivative of g(x) or the derivative
of u...that is going to be...we have the power rule so that is going to be 3 times 2 is 6x
minus 4. Voila, it is what we got when we used the Product Rule. So that sort of a validation,
not a proof, but it is validation that we have the same derivative function twice. We
got the same answer, so hey!!! This must work and look at how much less work that is:) The
Chain Rule is very very often is going to be a huge time saver and sometimes it won't
necessarily be. But it will a lot of times keep the notation a little bit simpler. We
are going to do a lot of examples in this video to make sure that you can do this yourself.
BAM!!! On that last example that I just went through follows what is called the General
Power Rule. It says if u is equal to a function inside a power where y 'u' is a differentiable
function of x and n is a rational number...so it can't be like the square root of seven
or something like that...then the derivative of y with respect to x is equal to n times
u(x) to the n-1 power. Just taking that power out front and reducing this by one and then
multiply by the derivative of function u with respect to x. Or, in simpler terms the derivative
with respect to x of u to the n is equal to n time u to the n-1 power times 'u' prime
or the derivative of u. This is a special case of the chain rule where you have power.
It follows this nice and easy pattern. We didn't really need to know that to finish
that last problem. This should be kind of ingrained in you thinking after using that
power rule. It is just the Power Rule and the Chain Rule combined together. So here
we have f(t) is equal to the square root of 4-t. Well again, we don't really like working
with radicals so I am going to rewrite this with a fractional exponent. That is going
to be equal to 4-t raise the the first power which you don't really need to write, but
the root is the second root. So square root, so we do need that exponent and it is equal
to one-half. We have 4-t to the 1/2 power. So, f prime of t is equal to 1/2...we are
going to bring that out front...so 1/2 times 4-t. This needs to get reduced by a value
of one, so 1/2 minus 1 is negative 1/2. We do not want to leave negative exponents in
our final answer. We are going to take this binomial that has that negative power and
drop it down into the base of the fraction now. Along with or in the denominator we already
have a 2. So we are going to have 1...that is that 1 here...one over, there is already
a 2 in the denominator see because this is 1/2 times this. So, this is really....kind
of like...well it is... over one. That would be times one, but that is not any more because
that 4-t is going to get moved down to get rid of that negative exponent. It is negative
1 over 2 times 4-t to the 1/2 power. It is perfectly acceptable to leave it in that format.
Or if you want to put the radical back in because that is how the problem originally
started, that is going to be negative one over 2 times the square root of 4-t. Next
example!!! For our next two examples we have f(x) is equal to 2 over 3 times the square
root of 2x-1. We are immediately going to change this to 2 over three times 2x minus
1 raised the 1/2 power. Now I don't want to leave this question in this format because
I don't want to use the quotient rule. What I am going to do is, because I have this 2x-1
in the denominator, I am going to bring this binomial up into the numerator and write it
as 2/3 times .... I have some room here. 2/3 times and I am going to let that 2x-1 come
up, and now I am going to have that exponent of 1/2 become negative 1/2. We kind of have
again yet another step more difficult. Another example of using that General Power Rule.
So, we are going to... Now this in here if you are doing your substitution, this is our
'u' ok. This is our inside function. I am going to apply the Power Rule. We are going
to do negative 1/2 times 2/3 which is going to be negative 1/3. f prime of x is equal
to -1/3 times our inside function of 2x-1. We are going to let that exponent be reduced
by one so it becomes negative 1/2 minus one is negative 1/2 minus 2/2 so it will be negative
3/2. Then we need to do that derivative, take the derivative of this inside function. That
is going to be the derivative of 2x-1 is going to be equal to 2. This is the u and this the
derivative of u. Ok. Well, this 2x-1 has a negative exponent so I am going to let it
float down to the base and get the 2 is just sitting there and the 2 is over 1. This is
going to become -1 times 2 is -2 over 3 and this binomial with the negative exponent is
going to get floated down giving us 2x-1 to the now positive 3/2 power. That can be rewritten
if you like as f'(x) is equal to -2 over 3...that is a power and square root... so 3 times the
square root of 2x-1. I don't want to publish something that has a mistake, so let me double
check my answer. We have a power of 3, and we are good. That is the derivative. Now over
here we have what will be eventually a very simple situation. I do not want to handle
this fraction in this fashion. I can very very easily, because this just a one, I can
rewrite this so that I don't have to apply the quotient rule as it sits right now. I
am going to take that base, that binomial in the denominator of 2t + 3 and bring up
into the numerator. That is going to be (2t+3) and now that whole binomial that is in the
denominator of the fraction... and fraction bars do act as a grouping symbol so it is
a binomial in the denominator. As I float that entire denominator up that has an exponent
of one, or you can show that it has an exponent of one, that will now will become 2t plus
3 to the negative one power. Then the original function of course was raised to the second
power. Now I have something raised to a power to another power. This is the Power to Power.
When you have stacked exponents, you should be multiplying them. This is going to become
f(x) is equal to 2t plus 3 to the negative two power. Now this problem is starting to
look like all the other ones did. I am going to apply the General Power Rule which means
that f prime of x is negative 2 times 2t plus 3, reduce that exponent by one so it would
be negative 3 now. This is my inside function, this is something raised to a negative two.
Now that inside function, that g(x) or that u has not been... We have not taken the derivative
of that yet. That needs to be found, that derivative of 2t+3 is going to be equal to
2. Again this is my u inside here and this is my u prime if you want to do those substitutions.
All righty then! The only thing with the negative exponent is the 2t+3 so that is going to get
moved down. This is negative 2 over 1, this is 2 over 1. Just this base with the negative
exponent is going to come down to the denominator. We get negative two times two is negative
four over (2t+3) now raised to the third power. Double check it, and we are good:) I have
got two more examples. They are quite advanced. I think what I am going to do, let's stop
this tape here...video...incase you don't need to see anything more difficult. The next
video with two examples will involve the combination of the Chain Rule and the Quotient Rule and
maybe a little bit of nested Chain Rule where we have to go through that process a couple
of times. If you are good here, thank you for watching. If not, the next video will
have two more advanced examples if you need them. I am Mr. Tarrou. BAM!!! Go Do Your Homework:D