BAM!!! Mr. Tarrou. In this lesson we are going to move on to from the Product Rule and work
out a few examples of the Quotient Rule. The quotient of f and g or f/g of two differentiable
functions f and g is itself differentiable at all values of x for which g(x) is not equal
to zero. The formula or the rule is the derivative with respect of x of f(x)/g(x) is equal to
g(x) times the derivative of f(x)... or in other words the denominator times the derivative
of the numerator minus the derivative of the denominator times the numerator over the denominator
squared. So we are going to take the denominator, put it first... take the derivative of the
numerator minus... and then the derivative of the denominator times the numerator. It
is very important that you get this order correct because in the middle here we have
subtraction. Subtraction is not commutative, so if you happen to get this backwards where
you take the derivative of the denominator first and then the numerator you are going
to get the wrong answer... where of course g(x) the function... we cannot get an answer
that is zero from the function because then we are dividing by zero and that is undefined.
So these examples are going to take up a little bit of room. So I need to erase this. I will
leave this up here so we can reference as we go through the work... but I need more
room than just the bottom of the chalk board to work these examples. So, I will be right
back. We have y equals 3x-1 divided by x^2+2. y prime, or the derivative of y, is going
to be equal to the denominator x^2+2 times the derivative with respect to x of the numerator
which is 3x-1... minus... now we are going to take the derivative of the denominator...
g'(x)... so it will be x^2+2 times the numerator... all over the denominator squared. Don't worry
about multiplying that out. Leave it as a binomial squared is perfectly fine. We have
x^2+2. The derivative of 3x-1 is equal to 3... minus the derivative of x^2+2. The Power
Rule. uh, I used the power rule here as well. We have 2x times... and the derivative of
a constant is zero so just 2x... times 3x-1... all over that binomial squared. All we have
to do now is clean up a little bit. Don't FOIL that out like I said, or distribute it
out. But we have 3 times x^2 and 3 times 2. We have... I am going to do this whole -2x
and distribute that through the parenthesis... We get 3x^2+6, and -2x times positive 3x is
-6x, negative time negative is positive so plus 2x, Oops excuse me x times x is x squared.
This all over x^2+2 squared. We have some like terms. We have 3 minus 6 is of course
negative three so -3x^2... plus 2x... plus 6 all over x^2+2 squared. And I always check
my work even though I made up these problems so just to make sure.. -3x^2+2x+6 over x^2+2
squared. GREAT. Next example. Whoo.. Next example. f(x) is equal to 2x+1 over the square
root of x. Ok again, I like to get those radicals out of there and write with fractional exponents.
Secondly, I discuss this with the Power Rule as well, these are long and tedious. I mean
you have to use the Power Rule or the Quotient Rule if you need to, but if you can adjust
the equation a little bit... the function you are given... to allow you to use the basic
differentiation rules you are going to save yourself a lot of work. So... And I would
be doing the same thing if this were just a simple constant of like 5, as well as this
square root of x. Because it is a single term and it will cancel nicely with the 2 terms
in the numerator. If I pull that fraction apart, when we add or subtract fractions...
when we have a common denominator we just copy that denominator. So we are going to
be able to pull this apart to be two separate fractions. So we are going to write this as
2x+1 over x to the 1/2 power. Then we are going to pull those two terms apart, in the
numerator, and write two separate fractions with the common denominator of x to the 1/2
power. That is going to be equal to 2x over x^1/2 plus 1 over x^1/2. And... well here
are dividing like bases and when you divide like bases you subtract the exponents. 1 minus
1/2 is 1/2, so this is equal to 2x^1/2. And though I am teaching the quotient rule I don't
want to leave this as a fraction. Because I can make this one with just the basic rule
of differentiation. That means I am going to write x to the negative 1/2. See now I
have got a nice... just non-quotient. So I can use the basic differentiation rules. So
f prime of x is equal to... I am going to take the derivative of each of these terms
separately and add them together. So we have... let's see here... I am going to take that
power of 1/2... Well, maybe I should show the work. We are just learning this right?
We are going to do 2 times the derivative with respect to x of x^1/2 plus the derivative
with respect to x of x^-1/2. Now I have pulled that constant out of the derivative using
the constant rule. So this is going to become 2 times 1/2x^-1/2. I am subtracting the power
by one, right? 1/2 minus 1 is -1/2. Now plus, pull this out front... I could just write
minus 1/2x, bringing that power down I need to reduce this by 1. -1/2 minus 1 is -1/2
minus 2/2, so -3/2. You know... 2 times 1/2 is just 1, so I am going to erase that and
just write 1. Ok. Well great! But you know I am still not really happy with negative
exponents in my answer. So I want to see if there is a way that I can rewrite this so
that I don't have negative exponents. I am going to do that with factoring. Now each
of these terms have an x in them. That means that I can pull an x out. But how many can
I take out? If you have not done this yet or you forgot from PreCalculus... If I say
we have 2x^4+3x^3, how many x's can I factor out of those two terms? I can factor out as
many x's as they can all give up. I can factor out three of them. I cannot factor 4 out because
this term does not have that meany. I look for the lowest exponent and that is how many
I can divide out. This becomes of course x^3 times 2x+3. So when you factor out a common
factor with variables of course you are looking for the lowest exponent. Well between -1/2
and -3/2, which one is the lowest value? The lowest value is the negative three-halves.
So we are going to do x^-3/2, and bare with me a bit because I am going to do this verbally...
no I am not... x^-1/2. What do you when you factor out? You divide it out. I just want
to point out that when you divide like bases you subtract the exponents. So from each of
these we are going to divide out x^-3/2. So when you divide like bases you subtract the
exponents. So -1/2 minus a negative 3/2 is -1/2 plus 3/2. Well that is equal to 2/2 or
1. So we have x^-3/2... this will become like I just said x to the first and we really don't
need to write that... minus 1/2... well I have x^-3/2 and that is exactly what I am
dividing by so that cancels and we just have x-1/2 inside. Well I still have a negative
exponent. Now it is just in one spot... in one factor. I can move that base with the
negative exponent down into the denominator and create a fraction. This is the shorter
of the two. If I had left this in quotient form I would be doing more work by the way.
So f'(x) is equal to x-1/2 over x^3/2. And we can call that a day, and be happy with
that. You could find common denominators and write this as what... 2x-1 over 2 and... That
is good enough. We have done enough work. It is a good place to stop for our derivative.
Again unfortunately with these derivatives, you might be happy with where you have stopped
and then gone to the back of the book and looked at their version of the derivative
and it could appear to be different because they simplified to a different point than
you did. So make sure that you continue to simplify to a point that your teacher is happy
with what you have got. I am just pausing to look at my own notes and that is not what
I wrote in my final answer. So let's make this look a little bit different. What if
I don't want a fraction over a fraction? Well, I could find that common denominator and then
flip the bottom up. But I think it would be easier if I make it look this way. If I go
ahead here and find a common denominator with these two terms and multiply this term on
the top and bottom by 2, then I have x^-3/2 times 2x-1 over my common denominator of 2.
Now when I go ahead and move that base with the negative exponent to the bottom I will
have a simpler version of the answer. I have 2x-1 over 2x^3/2. Now lets go ahead and call
it a day with this example and move on to the next example. For our next example we
have... If I left it in this form it would be lot of work to find the derivative of this
function. So I am going to take the x to the third power and distribute it through the
parenthesis and get f(x) is equal to... well if I have x^3 and I divide by x I have x squared...
minus 2x^3 over x+2. Now at this point you can find common denominators and put this
together, or you can leave this as two separate terms as I am going to do. Because we know
that when we find the derivative of two functions subtracted together I can find the derivative
of each individual term or function and then just leave it in that subtraction form or
simplify if needed. So f prime of x... or the derivative of x... that is nice and simple.
It is just going to be 2x.... and then minus with a set of parenthesis. Now leaving these
in two separate terms, I do need to be careful about that minus sign and make sure that I
distribute it through my parenthesis when I am done. So maybe you may feel it is better
to have this as a single fraction before you start the problem. But we are not. So we are
going to do the denominator, which is going to be x+2,... (Now I am going to kind of quicken
this up a little bit.) times the derivative of the numerator. The derivative of 2x^3 is
going to be 6x^2... minus the derivative of the denominator now because I took the derivative
of the numerator.... haha... the derivative of the denominator is going to be the derivative
of x+2, it is just going to be... that is a constant so the derivative is 0... and the
derivative of x with respect to x is simply 1... now times.... the derivative of the denominator
times (where am I at) 2x to the third. All of that is over the denominator squared, or
(x+2)^2. Do a little bit of cleaning up and we will be done. So we have 2x minus... 6x^2
times x is 6x^3... and 6x^2 times 2 is positive 12x^2... minus 2x^3... all over (x+2)^2. Let's
see. We have... Yes we do have like terms with the two cubed, or two terms with the
degree of three in there so we have 2x minus x minus... 6 minus 2 is 4x^3... plus 12x^2...
all over (x+2)^2. We could factor out a 4. That looks like a plus sign. It is supposed
to be a plus sign. I forgot the 1... plus 12. There we go. Now we could factor out a
4x^2 out of the numerator but it will not cancel with the denominator. We could find
a common denominator here and write one big fraction if we like. But we are going to go
ahead and quit with this and say that is the derivative of f(x) BAM!!! Last example coming
up. Last example. f(x) is equal to 2-1/x all over x+3. That would be a heaping full of
mess if you try to find the derivative of that so we are going to rewrite the function
a little bit, make it a little bit cleaner, and then go from there. I want to put those
two terms in the numerator together so I need common denominators. I am going to multiply
that by x on the top and bottom. We get 2x-1 over my common denominator of x. That is all
over x+3. Now you can multiply the top by x/1 and the denominator by x/1 to cancel out
that denominator of x, or you can do what I like to do which is any time you have a
fraction over a fraction you flip that denominator up and turn that division into the multiplication
of the reciprocal. So we are going to take that division again, numerator divided by
denominator, or fraction divided by a fraction and we are going to change that division to
multiplication of the reciprocal. We get 2x-1 all over x... let's go ahead and distribute
that together... x times x is x^2... and x times 3 is 3x. Right? yes. So now we have
f(x) is equal to 2x-1 over x^2+3x. That is almost exactly what I used as my first example
in this video. So I am going to walk out, give you a second to pause the video, work
it out for yourself, and then when we come back I am already going to have the solution
worked out. So I will break up the video and give you a chance to make sure you completely
understand how to do these problems and then we will be done. I hope you got it. We have
f(x) is equal to blah blah blah.... blah blah. I hope you got it right. I am Mr. Tarrou.
BAM!!! Go do some homework:D