BAM!!! Mr. Tarrou. In this video we are going to continue our introduction of Implicit Differentiation.
But unlike my previous video where I spend a lot of time showing you the comparison between
explicit and implicit forms of differentiation and why you want to at times prefer and do
implicit differentiation, in this video we are just going to do three examples building
up difficulty as we go so hopefully it helps you do your homework. We 3x^3-x^2y+y^2 equals
4. So we are going to take the derivative of both sides of this equation with respect
to x. That means that whenever there is an x variable we are going to find the derivative
and it will seem like normal, and everywhere there is a y we have to chain rule and remember
that is... Well we will see it. 3x^3, so that is... you know... differentiating with respect
to x so this is just going to be 3 times 3 is 9x... bringing the exponent down by 1 and
we get 9x^2... minus... now this x squared times y, that is a product of two factors.
So we have to use the Product Rule. That is going to be parenthesis... I am going to a
little bit less notation than I used in my past video here. The first factor times the
derivative of the second factor, so we have x^2 times the derivative of y. Well that is
going to be y prime, or dy over dx. ok. I am doing y' for shorthand notation. Plus...
let's see... Let's back there again. First times the derivative of the second plus the
derivative of the first... so the derivative of x^2 is 2x times the second factor of y...
plus... going along with the terms here. We have y^2, so that is going to be Chain Rule
or General Power Rule, 2y and bring that power down... so 2y with exponent of 1, we reduce
the power by 1. But now... That looks like a y prime so I am going to take that out of
there. So that is the derivative of the outside function. Again this is Chain Rule because
we are doing the derivative of y with respect to x. So the derivative y with respect to
x does not just cancel away. That is again dy/dx, or y'. This equals... Now this is a
constant at the end and when you find the derivative of a constant you get zero of course.
We are going to take the negative and distribute it through the parentheses and get 9x^2-x^2y'-2xy+2xy'=0.
We want to get all of the terms with y' or dy/dx alone, everything else to the other
side. So we are going to get... Well let's see here. -x^2y'+2yy'. These two terms that
don't have y' or dy/dx need to go to the other side. We are going to add 2xy to the left
and right, so plus 2xy. We are going to take this 9x^2 and bring it over to the other side,
so minus 9x^2. Each of these terms now both have a y'. We want to factor that out because
you cannot solve for a variable when it shows up in multiple places in the equation. So
we want to only be able to write y' once. We are going to factor that out and get y'
times (-x^2+2y) equals 2xy-9y^2. Now I got y' times something. I want to get y' alone
so I am going to divide both sides by this parenthesis... this grouping. We get y' is
equal to 2xy-9x^2 all over -x^2+2y. If it really bothered me I could divide -1 out of
the numerator and denominator to change the order around and have my squared term have
a positive coefficient. But me personally, I am happy with that. So let's get on to the
next example right now. BAM! What's up. I got a sine for you. 3sin(2x)-cos(4y)=1. Let's
get this differentiated [whispers] implicitly. Alright. So we got... Anyway whatever... 3...
the derivative of sine is cosine, so 3 times the cosine of 2x... and again I like to highlight
my angle measurements with parenthesis to make sure we understand it is not a multiplication.
So we have the derivative of sine is cosine, and that is the derivative of the outside
function. We have an inside function of 2x, so now this is my u. The derivative of u is
going to be the derivative of 2x is equal to 2. So 3cos(2x) times 2. Yes. Minus... and
I am going to do some parenthesis just to make sure we don't mess up our sings. The
derivative of cosine is negative sine. So -sin(4y). Now my inside function is u which
is 4y. So the derivative of 4y is equal to 4... and I don't why I wrote that 1 there
because the derivative of a constant is equal to zero. Ok. So... What have I forgotten?
Did you catch my mistake? This is ok for the 2x, it is ok to write 2 because I am doing
the derivative with respect to x. But here the derivative of y with respect to x does
not just disappear so I need a y' there. You almost let me forget that:) Ok. So what do
we have. Let's see here. The two negatives are going to cancel out right here. So that
can just be taken out and put in a plus sign. We are going to subtract 3cos(2x), or actually
6. Let's just clean that up a little bit. We have 6cos(2x)+4y'sin(4y) equals 0. We are
going to subtract 6cos(2x) from both sides and get 4y'sin(4y) equals -6cos(2x). We need
to get y prime by itself. Well y' is being multiplied 4 and it is being multiplied by
sin(4y), so we are going to undo that multiplication with division. We get y' is equal to -6cos(2x)
all over 4sin(4y). Oops. Yes... Excuse me... over 4sin(4y). Now the 6 and the 4 will reduce.
They are both even. So 6 divided by 2 is 3 and 4 divided by 2 is 2. So we get -3 over
2. You cannot do that reduction though with the 2 and the 4 because those are inside of
the cosine and sine functions. Again it is not just cosine times 2 times x and sine times
4 times y. It is the cosine of 2x and sine of 4y and thus there is no more reduction
possible. I am just making sure I did make another silly mistake. Got all my y prime
marks in the right spot. I am Mr. Tarrou. Let's get to that last example. For our last
example. I have erased my title and given myself more room. You know what that means.
It is going to take some work! We have the square root of xy minus 2x^3y^2 equals 1.
We want to find the derivative of this implicitly. And even that way it is going to take some
work. We are going to do a couple of things to make sure that I don't make any careless
mistakes as I make this video. I am going to write xy to the 1/2 power. Again I never
like work with those radicals. I want to just emphasis that... When I find the derivative
of the second term, there are two factors in here. Actually there are three but that
is just a constant. So I am worried about the x^3 times y^2. When you use the Product
Rule you get two terms and I want to make sure that we get this signs correct and you
will distributing that negative sign through. You could probably handle it with the first
factor be -2x^3, but I like to do all the little small steps. It helps me make sure
I get the right answer all the time. It will probably help you not make any sign mistakes
either. So let's walk through this. We will take care of that negative a little bit later
distributing it through the two terms we get from the Product Rule. We have the Power Rule
here. So we are going to do... and we are going to have to do the chain rule because
there is a product inside of here. We have 1/2 times xy. This is now going to get reduced
by 1. So this is going to be negative 1/2. We are going to have to take the derivative
and we have got a lot of room... and besides with this being a video I can bore you for
as long as you will watch:) Right? haha. ...and show you all the steps. The derivative with
respect to x of xy... then minus... let's go ahead and work this out here. We have got
a factor in here, so we are going to take... with the product rule... the first factor
times the derivative of the second factor. This is going to be first factor which is
2x^3 times the derivative of the second factor... and we are going to need the chain rule every
time we get to those y value because we are differentiating with respect to x and not
y. So 2y and that gets reduced by 1, but I am not going to write that because it looks
like a prime symbol... so we have 2y times the derivative of y with respect to x or just
y'. So that is first times derivative of the second plus the derivative of the first...
which is going to be 6x^2... take the power down and then reduce the power by one... I
don't need to write an x' because I am finding the derivative with respect to x. So I don't
need to have that extra x' in there (dx/dx). So the derivative of the first times the second,
so y^2. This equals... Now we are finding the derivative of both sides with respect
to x. This is a constant, so that is going to be zero. Now let's turn our attention to
this product here. At the same time I am going to get rid of this and write it as 1 over
2 times the square root of xy. This is the product again, so like it was here... first
times the derivative of the second which is going to be y' plus the derivative of the
first... now the derivative of x is 1... we know that by now... times y. I am also going
to take this negative and distribute it through the parenthesis. So we have -2x^3y^..... Let's
go ahead and to this as well... 2 times 2 is 4. So -4x^3yy'... negative and positive
is -6x^2y^2 equals 0. Let's do the distribution again to get that 1 over 2 times square root
of xy through the first set of parenthesis and we have xy' over 2 square root of xy plus
y over 2 square root of xy minus 4x^3yy' minus 6x^2y^2 equals 0. Ok. Whoooo! Now we need
to get the terms together that have the y' in them, and everything else to the other
side. So these y' are going to stay here. I am going to move these other two terms over
with addition and subtraction. We have xy' over 2 square root of xy minus 4x^3yy' and
that is over one but we are not going to find common denominators... but whatever. Let's
do the negative first because that is going to be positive, so take this -6x^2y^2 and
move it over with addition. We have 6x^2y. This positive term is getting moved over with
subtraction, so minus y over 2 square root of xy. Ok. Well you can't solve for a variable
again if it written multiple times. I want y prime. So I need to factor this y' out so
that i can get it written once and eventually get something to where it is y' equals everything
else. So the y' is going to come out and we get y prime times x over 2 square root of xy minus 4x^3y equals
6x^2y... I got a little sloppy with my handwriting here didn't I... What is that? That is supposed
to by a y... minus y over 2 square root of xy. Ok. Well I am not done until I get y'
alone so I need to divide both sides of all of this and we get y' is equal to 6x^2y minus
y over 2 square root of xy all over x over 2 square root of xy minus 4x^3y. Ok. Well
that is great. Let me make sure before I keep talking that I have not made a careless mistake
anywhere. Oh oh. y squared. y squared. y squared. Talking and writing and thinking is sometimes
is beyond my capabilities. Ok. Good. I am glad I caught that! Now. This is a, it is
going to be a complicated fraction anyway. We have fractions over fractions. So... um...
Let's clean this up a little bit. Let's multiply the numerator and denominator by 2 square
root of xy. I am multiplying by one, so I am only going to make it look different. Just
to clean up these denominators. So we have y'... almost done here. 6 times 2 is 12, so
we have 12x^2y^2sqrt(xy) minus... so many places to make a small mistake... we are multiplying
by two. This is getting again distributed. This is why you want to be very careful with
your notation, your work in Calculus. We have a 2 and a denominator factor of 2, so they
are going to cancel out. Actually this is all going to cancel out, so we have minus
y... all over... When we distribute here, the denominator is once again going to cancel
out and we have just x minus... 4 times 2 is 8x^3ysqrt(xy). Before I call this a day
let me make sure I have this correct. 12 x squared.... [mumbling] Good. Ok! So our last
example is done. I hope these three help you to do your homework and better finish your
questions involving implicit differentiation. I am Mr. Tarrou. BAM!!! Go Do Your Homework:)