Implicit Differentiation Explained: From Concepts to Applications

Name: Calculus 1 Lecture 2.7: Implicit Differentiation
Uploaded: 2026-03-19T09:49:29.327073+00:00
Duration: 1 h 8 min 10 s
Channel: Professor Leonard
Description: Summary and key takeaways on Calculus 1 Lecture 2.7: Implicit Differentiation — Summary, covering to Explicit and Implicit Functions An explicit function

Professor Leonard

Mar 19, 2026

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68 min video

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An explicit function isolates the dependent variable y on one side of the equation, for example y = 3x + 4. In an implicit function the variables x and y are intertwined, such as x² + y² = 4, and y is not solved for directly. Some implicit equations can be rearranged into explicit form, but many define more than one function; the circle x² + y² = 4 represents both an upper and a lower semicircle. When an implicit relation cannot be easily solved for y, the derivative must be obtained without an explicit formula.

The Need for Implicit Differentiation

Explicit differentiation relies on having y expressed as a function of x. If solving the implicit equation for y is difficult or impossible, that approach fails. Implicit differentiation provides a way to compute dy/dx directly by differentiating the original relation, sidestepping the need to isolate y.

The Process of Implicit Differentiation

Differentiate both sides of the given equation with respect to the independent variable x.
Treat y as a function of x (y = f(x)). Whenever a term containing y is differentiated, apply the chain rule, which introduces a factor of dy/dx. For instance, d/dx(y³) = 3y²·dy/dx.
After differentiation, collect all terms that contain dy/dx on one side, factor dy/dx out, and solve algebraically for dy/dx.

Mechanisms & Explanations

Treating y as a function of x ensures that any operation on y reflects its dependence on x.
Chain rule for y works exactly as it does for any composite function: differentiate the outer part, then multiply by the derivative of the inner part (dy/dx).
The derivative of x with respect to x is 1, so pure‑x terms contribute no dy/dx factor.
Solving for dy/dx involves moving all dy/dx terms to one side, factoring, and dividing by the remaining coefficient.
Second derivatives are obtained by differentiating the first derivative expression, often requiring the quotient rule and substitution of the previously found dy/dx.
Tangent line equation follows from the slope m = dy/dx at a specific point (x₁, y₁); the line is then y − y₁ = m(x − x₁).

Examples and Applications

Polynomial implicit equations such as x³ + y³ = 6 can be differentiated term‑by‑term, applying the chain rule to the y‑term.
Trigonometric implicit equations like sin(x) + cos(y) = 0 require the derivatives of sin x (cos x) and cos y (‑sin y·dy/dx).
Second derivatives appear when curvature or acceleration on an implicitly defined curve is needed.
Slopes of curves at particular points are found by substituting the point’s coordinates into the solved dy/dx expression.
Tangent lines are constructed using the point‑slope form after the slope is known.
Related rates problems use implicit differentiation to relate the rates of change of two or more quantities that satisfy an implicit relation.

Key Concepts and Rules

Product Rule and Quotient Rule are applied whenever products or quotients of functions involving y appear.
Chain Rule (General Power Rule) underlies every differentiation of a term containing y, ensuring the factor dy/dx is included.
Fundamental derivative facts remain unchanged: the derivative of a constant is 0; d/dx(xⁿ) = n xⁿ⁻¹; d/dx(sin y) = cos y·dy/dx; d/dx(cos y) = ‑sin y·dy/dx; d/dx(ln y) = (1/y)·dy/dx; d/dx(eʸ) = eʸ·dy/dx.

Takeaways

Implicit differentiation finds dy/dx without solving the equation for y, which is useful when the implicit relation is difficult or impossible to isolate.
The method works by differentiating both sides with respect to x, treating y as a function of x and applying the chain rule whenever y appears.
After differentiation, algebraic manipulation isolates dy/dx by gathering all dy/dx terms on one side and factoring them out.
The technique extends to second derivatives, product, quotient, and chain rules, enabling slope calculations, tangent line equations, and related rates solutions.
Implicit equations may represent multiple functions, such as the upper and lower halves of a circle, so the derivative corresponds to the specific branch at the point of evaluation.

Frequently Asked Questions

Why is implicit differentiation needed when solving for y is difficult?

Implicit differentiation is needed because explicit differentiation requires y expressed as a function of x; when the equation mixes x and y and cannot be solved for y, differentiating implicitly lets us compute dy/dx directly by differentiating both sides and using the chain rule, bypassing the need for an explicit formula.

How does the chain rule appear in the derivative of y³ during implicit differentiation?

When differentiating y³, treat y as u(x); the derivative follows the power rule d/dx(u³)=3u²·du/dx, so d/dx(y³)=3y²·dy/dx. This extra factor dy/dx is the manifestation of the chain rule, ensuring the derivative accounts for y’s dependence on x.

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Full Transcript YouTube

so today you're going to learn something
called implicit differentiation you see
so far what we've done is worked with is
explicit forms of equations let me give
an example this right here when you have
y equals that function in terms of X or
you have one y its salt or equals some
function of x such as y equals 3x plus
or is considered an or maybe the 3x
squared plus four is an explicit form of
a function explicit means lies over here
your X is over here you can solve for y
does that make sense for you this is
explicit means Y is given explicitly
tells you exactly what Y it is on the
other hand
is this an explicit function is why
solve for itself and then X's on the
other side no as a matter of fact you
see Y's mixed in with X's this is called
implicit form implicit means yes there's
a y there and you're going to say that Y
is a function of X but it's not written
as y equals so Y is still function of X
it's just bound up in the problem that's
what implicit means so this is implicit
that an interesting thing is a lot of
time not a lot of time some of the times
we can make an implicit function into an
explicit let me show you how in this
example first thing we want to solve
this for y I would probably factor out
the Y I get 1 plus x equals x are you
with me on that and if I divide by 1
plus X I get egg y equals x over 1 plus
X and that's now explicitly written so
sometimes I can change from implicit to
explicit so if the Y is alone on one
side that means it's written explicitly
so explicit
why alone on one side
how about this wanna see if you kind of
get the idea
would you say this is implicit or
explicit listen explicit next course it
would have y equals implicit would have
y's and x's all mixed up what's this one
definitely could you solve it for y okay
let's try we solve for y we're going to
get y squared equals four minus x
squared and how would you get Y that so
so Y would equal the square root of four
minus x squared true or false
think carefully three or four what's
wrong very good because when you take a
square root you have to put the plus and
minus right so here's what this says
well actually what is this do you
recognize the square root of 4 minus x
squared and the negative square root of
4 minus x squared what are those know
what I mean what shape on the graph is
that what's this that's a half circle
this is a lower half circle do you see
this this is our two our upper that's a
circle by the way radius of two centered
at zero zero so this is an upper half
circle and a lower half circle what that
means is that implicit functions can
often define two different functions our
sorry implicit equations can often
define two different functions this is
not a function by itself because when I
plug in one number I get out two answers
that says it's not a function it's an
implicit equation it's actually two
different two distinct functions an
upper and a lower half circle do you see
what I'm talking about so oftentimes
that's what happens implicit equations
can define one more than one function of
X
just like I did here this is implicit
that's two different functions so so far
we know what the explicit means I hope
we know what implicit means I hope that
means bound up within that why still a
function of X sometimes we can solve for
it and change it from implicit explicit
and sometimes an implicit equation will
make more than one function of X now I
have to consider this example how about
this
would you say implicit or explicit what
do you think super input was another
level super implicit it's like super
whoa huh you agree that is implicit
right okay good you understand that
could you solve it for y could you solve
her for y in such a way that's why by
itself equals a function completely an X
with no other X's I mean sure you could
divide by 9x but that's still a y let me
refresh it or what explicit means
explicit means there's one wide
it's a lone on one side everything else
is a Texas can you do that here yeah try
it try it if you want
can you do it easy I love Mexico she can
do it easily here no no you can't
because the way you might do is to
subtract this over subtract this over
right but you need have 3y cubed minus
9x y you'd factor out a Y but you still
left with Y squared in there right
there's really no good way to get rid of
that at least not easily so can you
easily get this into an implicit I saw
right in explicit form can you do it as
there's no this this cannot be written
explicitly if there is a way it's not
going to be easy if that sometimes there
is no way you can't write it explicitly
for some equations now to keep in mind
what we're doing here we're not doing
basic algebra we're doing calculus and
calculus involves derivatives right yes
okay no we've been doing nothing that's
hold on
derivatives so here's a deal if this is
implicit and I can solve for explicit I
could take the derivative of that in
fact that would involve a quotient rule
right and this one I could actually take
the derivative of the top function or
the bottom function and be just find
I've involved a chain rule or the
general power rule because that one-half
this one though there's actually no way
that we're going to be able to get that
as y equals something in terms of X with
me what that means is even if you got Y
by itself there's still going to be
otherwise and here they're going to mess
you up we don't want that so we have to
have another
method to find derivatives instead of
just having y equals a function and X
there's got to be a different way and
there is and it's called implicit
differentiation that's what we're doing
today so we're going to talk about how
we take a derivative of this thing
without having to solve it for y now
that's kind of interesting question
right can you take a derivative without
having at y equals yeah you can we're
going to look at that today so for
example it's not the calculus isn't hard
the algebra might be a little bit but
the calculus is not hard implicit
explicit everybody what is that explicit
would be y equals something is that the
case
this is explicit this is implicit this
is implicit this is implicit unless it
says y equals then it's not explicit
what does this one again it's definitely
implicit now we could actually solve
this one for y do you see that we'd
subtract x cubed we take a cube root it
would be just fine you can find the
derivative that way
but what I'm trying to show you today is
how to do this without having to do that
by just solving this derivative or
solving for the derivative without
having to solve for one because of
examples like that which we're going to
do that so we could solve for y take
derivative or we could use implicit
differentiation here's the whole deal
though in order to do this you must
treat Y as a function of X because if
you could solve it for y y would equal
something in terms of X right so you
have to treat it and that's going to be
important here in just a second you see
why in a bit you must treat Y
as a function of X
must realize function X
you know how you can add something to do
such an equation subtract something both
sides equation multiplying by take
exponent take a root pretty much do
almost anything you want to to both
sides equation as long as you do to both
sides you know that right
did you know you also take a derivative
at both sides of the equation did you
you do now don't you do anything to
equations you want to pretty much a long
long with taking a derivative you can do
that so I could actually take a
derivative with respect to my variable
what's my variable that I'm dealing with
here my there's two variables so how do
you how you tell which one we're talking
about Y as a function of X right so you
can have this with respect to X so we're
still treating X as our variable Y is a
function of that variable that you just
don't know explicitly you don't know
exactly what it is you get the idea so Y
is good y equals something we don't know
exactly what it is because we don't want
to solve for it we don't really care
that much like this one you don't care
that much you're saying okay I know Y is
a bunch in terms of X and that's what
we're dealing with here so we're going
to take a derivative of X cubed plus y
cubed and that equals what do you think
5 or the derivative of 5 derivative of
both sides so we treat Y as a function
of X that we simply just don't know we
don't know what Y actually equals and we
don't really care
here's our steps then I've shown you the
first one the first one is you're going
to take a derivative or differentiate
both sides with respect to X
just make sure your dutiful decides it's
an equation after all you have to do
that so drive both sides
let's check the X I'll start you off
with the easy side the easy side is the
five side let's start the five side can
you tell me what's the derivative of
five which is a constant very good if
you leave it at five you see that would
really affect your problem that would be
a problem
zero now let's look at here can we take
the derivative of these pieces they're
added together which means I can take
them term by term you with doing this
I'm glad you're here today this is
something you really need to see now the
derivative of x cubed is pretty easy to
do we're taking a derivative with
respect to X you fall down that in fact
if I wanted to since that separable by
addition I could do I'm going to show
this to you one time we typically won't
do this but I show to you here you could
write x cubed derivative X cubed plus
the derivative DDX of Y cubed okay with
that so far right yes ma'am what's the
derivative of x cubed with respect to X
how much is that
makes for a very good plus now this is
the weird one
what's our variable that we're taking
the derivative with respect to two okay
we're taking with respect to it says
right here fun is that
put that it's a Y now here's the idea y
is a function in terms of X right so
basically this is interesting I hope you
find this interesting so you're saying
what you have here is actually a general
power rule you actually have that do you
see it
it really is this it's really that it'd
be the same thing
check this out it'd be the same thing as
doing this 3x squared minus 1 to the
fourth is this a function of X yes it's
being raised to some power right and how
you do this is actually the chain rule
the special version the corollary which
is called the general power rule you
with me on this you bring down the 4
watch carefully I'm gonna draw the
analogy right here you bring down the
form if you find that it found the
derivative you'd have one out 2/3 you
groove this but then you don't end there
do you
I made it
that wasn't evil like yet wait
I have good ones I thought so easy we
bring it down because that's a general
power we'll bring down the exponent fact
one from it derivative of the inside
your this right you did homework on that
I hope if you didn't well here's what
you were supposed to do that we go
that's it and you would get whatever the
derivatives of that and you'd go 6x for
that derivative right now
look at the comparison this is a
function in terms of X that was a
function in terms of X this is raised to
some power that was racism power so the
derivative says this it says sure you
bring down that exponent and subtract
one from it this is no different than
what you've been doing this is actually
this right here are you with me the
problem is this in this case you had to
multiply by the derivative of the inside
here do you see what I'm talking about
what's the derivative of what's the
inside here the Y so we multiply by D DX
of what are you clear on that do you see
the chain rule at work do you see that
you need to see this before you'll any
further raise your hand feel okay I'm
getting that part of it she was coming
from now here's the weird thing D DX of
Y that says that says this check it out
this is the derivative of Y with respect
to X does not the derivative of Y with
respect X
this says the derivative of Y with
respect to X but then so does this the
derivative of Y with respect to X
doesn't that say the same thing here and
what does this represent that represents
your derivative you don't get it you
know the whole thing is no this is your
derivative you get this that's the
derivative of your Y every time you take
a derivative of Y we were going to get a
dy/dx because every time you take a
derivative of Y you actually have a
chain rule every time you have a chain
rule you take a derivative of the inside
of it does that make sense to you so
every time you're going to have a DDX of
why why this is the derivative of Y with
respect to X right this is the
derivative of Y with respect to X that
is your derivative actually if you think
about like variables it's kind of weird
but D DX times y over 1 dy over DX I
don't know if that really works but
they're smoother than that you can do Y
DX that's what these two things mean
they're equivalent statements equals
zero vision if you look at getting down
that far so here's the whole thing for
you you take it derivative of both sides
with respect to X since you're treating
Y as a function of X every time you take
a derivative of Y every time you must
get a dy/dx that's really the only
important note for today if everything
else which is say so each time you take
a derivative
every time you derive why you should
have dy DX on account of the chain rule
each time you drive why you must get a
dy/dx
on account of chain rule
now some people might be asking well mr.
Leonard why don't we have a 1 for this
why don't we have to use the chain rule
here well think about it think just
think for say if we did the same thing
you would do chain rule and you would
get d DX of X right if you did the same
thing that's DX over DX
how much is DX over DX 1 doesn't matter
right that's all we don't have to do
that so for our dy DX so that is
different this is not the same variable
is y this is a function in terms of X
which means you need the dy/dx there and
that's the whole crux of implicit
differentiation now the only issue is
this stands for a derivative this whole
thing is all mixed up together can you
solve for this
this was the calculus the algebra is the
rest of it get this by itself and then
you have your first derivative ok
subtract this what another is this
so you keep the keep the Y's the 3 y
squared well for right now because if
you multiply it
so I'd subtract now we divide
so now that's gone if we do just a
little bit of simplification we get
dy/dx equals looks like negative x
squared over Y squared me can you make
it down that far that's our derivative
notice how our derivatives also implicit
our function was implicit that makes
sense our derivative most of the times
will be implicit as well
now could you have done this and got the
same exact answer the answer is yeah but
check this out take yourself if you did
this explicitly so if you solve this for
Y which is that which is that if you
take the derivative of this thing are
you going to get exactly this no because
this won't have wiesen in this will
however you can make a substitution
how much is y equal to that you can
actually substitute in and that will be
the derivative there isn't that kind of
interesting and they kind of cool did
you see what I'm talking about take over
your head right now hope none less far
notice how if I were to solve it for y y
equals this right some that Y is here so
this would be
that you can do which is the derivative
if you would to if you do the strip
tremolo see if I'm right
pretty sure I am let's see if I am right
see if you can derive that cute and yeah
it's cute it's a cube root hey do that
get your to the negative three halves
power first our negative 2/3 powder
that's going to be right here for you
you're going to have your X squeezy x
squared in there right the 3 is going to
be gone because you have a 3 X square
negative 3 x squared you have over 1/3
you cross those down you'll have a
negative there's your negative you have
an x squared is your x squared you have
this thing showing up again to a weird
power it is the same derivative it's
just this one is not explicitly written
this one is with this function you could
do this so you won't be able to do this
with your with your derivative in most
cases because you can't solve for y so
that's the the beauty of implicit is
that this is just as valid as that is
just as valid so we do this step one I
used to learn one step always step to
solve for dy/dx so write a step two on
your paper somewhere
step two is solve for dy/dx
so to solve this okay guess what that's
it that's implicit differentiation if
you had this idea down you have the list
of differentiation down now is it going
to get tougher of course yeah because
we're going to incorporate things like
changeable things like the product rule
things like quotient rule what you can
all do it makes it Alex were nasty the
calculus is not more algebra is yes
difficult in some cases would you like
to try some examples to see this thought
you would okay let's take a look at just
few of them first one I'll start off
with is this one
I say find dy/dx so if we want to find
dy/dx that means of course we're
implicit right here there's no way we're
solving for y oh my gosh could you
imagine solve that for y not a chance
so when I can solve that for y that'd be
ridiculous good thing we have implicit
differentiation right now so Blissett
differentiation says all right why don't
you go ahead and take the derivative of
both sides I want to see your work on
the stuff show me your DD X's
just don't forget both sides don't
forget them
now let's look at the right hand side
that's going to be the easier to two in
this case what's DDX of 4x to the 5th
how much is that
and so don't forget the stuff you
already know how to do don't let that
leave you now this is where we get to be
kind of careful right if we have DDX of
3y squared plus sign 1 you know how to
take the derivative of this place you
know how to take the derivative where
you're dealing with what idiom of X's or
Y's hear what those are functions of X
which means that when you take a
derivative you will get a dy/dx are you
clear on that every time you touch a
piece of your function that has a y in
it with a derivative you get a du by DX
every time because of the chain rule I'm
gonna stop showing you this right here
this whole chain rule idea we're just
going to do things like this check it
out 3y squits a derivative of 3y squared
6 wide great 6y what did you take a
derivative what did you then do a dy/dx
the reason is again you would have a
chain rule chain rule would say bring
down the exponent multiply it take one
away from it that's the 6y with then x d
DX of Y right that's what we have over
here d DX of Y but that is your dy/dx
and that's where that's coming from
he'll be ok with 6 y dy DX if knows you
guys have questions on that right now
this part is coming from the same ideas
this right here it's the chain rule you
bring it down you subtract derivative of
the inside and that's what we're doing
derivative of the inside in this case is
d DX of Y that is dy DX that's that
piece you need that piece otherwise you
won't be able to solve for your
derivative you're with me gonna have
that so we need that Plus as we have a
plus here what's the derivative of sine
because I okay so we have cosine of Y dy
D dy DX
you took a derivative of something
involved Y that means two languages
again if you want to see this this would
be the chain rule you go cosine of Y
leave that leave the inside alone DDX of
the inside but that's just a wash
that's dy/dx why we're getting that
thing that's where that's coming from
do you see where it's coming from that's
a good question let's solve for this and
I'll ask you that I actually have a
couple examples later on that will
illustrate what you do for that so we're
getting there right now I just want to
get to handled on on the drill record of
stuff okay whoo oh oh man what do you
got to do now yeah you got to solve it
so when you solve for a variable when
you solve for variable you get all your
variables to one side right then you try
to combine like terms and you try to
factor until you can just one of them
and divide that's how you basically
solve for a variable or a variable here
is dy/dx so get all your dy/dx is to one
side bass girls with your calculus is
done you're done you've done the
calculus correctly
now it's algebra so you see your dy/dx
here your dy/dx here that means you're
going to factor so you're going to be
good at factoring if you factor that out
you get a 6y plus cosine y equals 20x -
2 4
so far so good let's get rid of this
nasty piece of junk what are we going to
do with that just to buy it this is
being multiplied correct so if we divide
both sides by that so divide here divide
here we're dividing by this expression
we get dy/dx equals 20x to the fourth
cool over this same expression 6y cosine
one
is it really too hard not really not too
bad they don't get harder when we start
doing some other rules in there which
are about to start right now and you see
that but in general they how this is
going to be very similar very similar
stuff you've done as long as you
remember what what's the important piece
of information here yet ever dy/dx
that's the important piece information
practical application do we have if
we're trying to find like the perfect
music with for a circle why why would
you want to find the top half turn of
the bumper circle and the box well if
you mean those things exist in your life
right so if we somehow need to find a
tangent line of those you can do it so
we can actually find tangent lines to
these weird curves which aren't even
functions not functions we can still
find rumors of that it's going to come
back at you later in calculus to for
when you have more than those variables
right so you're going to have use for
this later on as well translate that's
your fiction is it pretty something
where you have a multi variables and
you're holding one constant you trying
to solve you can you can actually do
partial derivatives things like that I
don't know if that's I think that's the
same thing isn't explicit if you use
that in that context and I'm sure but I
do know that this is useful for finding
those derivatives and the slopes of just
in fact if you really think about it
funny a trivet you want you've been
doing this whole time if I give you an
explicit derivative or X or some
function y equals rigged square any of
you sometimes if you take the derivative
of both sides which is actually what
you've been doing check it out you have
DDX of y equals d
of 3x squared minus 4 now this sense is
a function in terms of X or 6x this
that's a function in terms of X in it
derivative gives you dy DX you've
actually doing the whole time you've
been doing implicit it's just been
nicely solved for you the whole entire
time that's kind of neat right so is
their applications for implicit
differentiation yes because you've been
doing this entire time implicit just
comes about because sometimes you can't
solve for what and that's why we have it
still be the same thing still a
derivative just you cannot solve that
for one good question are you ready to
move on are there any other questions on
this before we do our ufi yet and then
kind of cool all right I think it's cool
no no that's later I'll get that for ya
no no that's this chapter coming up
don't worry
see look simple look simple example why
do I have a pulse huh implicit explicit
what was this definitely implicit
because the Y's not my self
can you take an implicit derivative of
this what is y a function of X what
happens when you take a derivative of X
which is a function of x times y which
is a function of X you see the product
rule kind of bug so what we can't do and
you just go oh X is one y is dy/dx da
have done because then dy/dx would equal
1 right and that's that's not the case
in fact if you ever solve this what
you're going to get is negative 1 over x
squared that's what that is because if
you divide by actually one of our
extruder that it's next to negative 1
power bring that down subtract one from
it you get negative 1 over x squared you
see that on top that's got to be the
same derivative so here when you have
the product rule yeah you got to use it
that's a function of x times a function
of X so sure we're going to take a
derivative of both sides and you let
your your DDX kind of do the work for
you just follow that down DDX of one
what what's D DX of 1 0 so don't forget
about that you always take the
derivative of both sides D DX of X times
y and that's a function of x times a
function of X and that implies the
product rule so we go okay let's do
derivative of the first times the second
plus the first times the derivative of
the second
once you help me out with this folks you
tell me what's the derivative of X with
respect to X good and what else do I
have
why do I need a dy/dx here did I take
the derivative of Y then no you don't
need it plus I know this X doesn't
change what's the derivative of Y with
respect to X yeah that's kind of nice
right
can you solve for dy/dx are you ok I'm
getting it for okay solve for dy/dx what
do we do subtract I so X dy/dx equals
negative 1
goodbye my ex because that's being
multiplied right now if you divide by X
you get dy/dx equals negative Y over X
now I made you the guarantee that this
is going to be 1 over negative 1 over x
squared and this is not negative 1 over
x squared this is negative y over x
squared why is that well because I solve
this derivative by using implicit if I
had made this exquisite which you could
do easily that's 1 X true white is 1
over X if I plug in one of Rex well I
get negative 1 over x over x over 1
flip it reciprocate you're going to get
negative 1 over x squared which is the
same exact derivative that's the same
thing now this is one of the rare cases
where you could make your implicit
function into explicit and do the same
way I'm just giving you this as a kind
of a for instance that when you see this
you need the product rule like can you
see the product rule here okay so do you
need to do this no no no you know this
is a valid answer when you're doing
implicit this is just fine this is what
I'm looking for as a matter of fact so
don't forget what the plug for you've
got to see the product rule
well I don't know if this is fun or not
it's fun for me I love this stuff but do
you understand it is really the main
thing it's what I care about whether you
find the fun honestly I don't care I
would hope you do but I care that you
get it do you get it after the practice
yeah oh be a boy yeah yeah I don't
expect you to get this out on the first
go I mean if you do great and you're
able to do all of them sure but yeah
there's going to be some tough ones in
there you are going to have to Frank I
also need to show you this on can you
could you could you take the derivative
of that hey why don't you practice that
right now go ahead by the derivative
find dy/dx
do you need a product rule for this
problem
thankfully I know
okay so that becomes 6x that becomes to
live did you get the dy/dx if you lose
the divide e^x you get lost on the
problem right because then you don't
have a derivative solve for and you're
like what am I supposed to do now is
make any sense at all so you have to
have those dy/dx is that stands for your
derivative did you make it that for
everybody helping show again something
way that for good okay now we solved for
solving Ford says well actually it
probably nice if I added this to both
sides and divided by 2y but I'm guessing
most of you subtracted 6x and divided by
negative 2y you still get the same
answer but if divided by negative
because that doesn't matter negative 6x
if we divide by negative 2y
you're going to get positive did you get
positive 3x over why yes everybody great
now unfortunately this is not the
question I'm asking you for that's the
question I'm asking for okay a second
derivative now can you take a second
derivative this thing of course you can
but in order to take a second derivative
yes you have to solve for the first
derivative so well to do it correctly
you have to solve for the first
derivative so we solved for first
derivative the second derivative just
says hey check this out this kind of
neat why don't you just take a
derivative of both sides so go ahead and
and do that can we take a derivative
both sides watch how how they accomplish
it if we derive both sides we get D DX
dy/dx equals D DX 3x or what can we
start with the easy side first first
easy side first
I took a derivative of the derivative
gives you a second derivative and you're
going to see the notation coming to play
do you see this like it
how many deeds you have to D so d
squared Y remember where that comes from
okay cool over the limit doesn't work is
this D don't get d squared x squared
really should if you wanted to do that
it should be like that we just never
show that this becomes d x squared and
that's where you get the notation for a
second derivative are you seeing where
that notation is kind of coming from
it's cool I guess if you're me it's cool
I don't get out much
what are you going to have to do for
this use a push rule go ahead set up the
quotient where you know the quotient
rule now go ahead and do it just
everywhere it's easy it's an easy
derivative just everywhere you see a
derivative of Y you have to put 2 uy DX
don't forget that and then you're fine
I'll make it sound so what you'll see
when I'm doing it I said then you're
fine
Oh before then yeah use the quotient
rule set that up everywhere where you
take a drinkable Y in the you dy/dx
what'd I tell you do low D high minus
high D low ok so low D high minus high D
low
again not to be confused with CeeLo blue
CeeLo is gathered glasses on the voice
very cool show my bedroom all over
bottom squared I guess I could use
parentheses there as well did you set up
the quotient rule correctly so that
stuff's not going to change I've taught
you I told you there's only three rules
I've touched all three rules now is
justify them so do I put it do I take
the derivative that Y so this is going
to be Y times do I need dy/dx for this
one for this one no because I'm not
taking different one that's going to be
three minus three x times what's this
going to come to my white and there it
is
perfect y squared that gives you if we
clean this up just a little bit 3y minus
3x now wait a second
this is weird because I have a
derivative and I have a group that's
strange
do I want to solve for that no no the
reason why you don't want to solve for
that look at the board
have you already sold for how much is
that equal to okay just plug it in so
you're going to find dy DX's but if you
take the second derivative you will have
already solve for dy/dx that's cool
now chances are for you guys you might
not always get it like this all right if
you if the back of the book looks
different than this what they probably
did because you have complex fractions
they got rid of that why if you want to
get rid of easily here instead find a
common denominator just multiply
everything by y and that to get rid of
it so multiply Y multiply both by Y
you're going to get 3y squared minus 9x
squared over Y key you see we're talking
about that should be the same thing as
many common
would you markusrohner the testicles you
know I'm not sure off to think on that I
would like it as simple as possible
now it's sometimes your any some nasty
things and would take you more time than
than that so either I won't give you
stuff it's going to be really nasty now
you sit like this that you can simplify
pretty easily or I would say just okay
now I'm going to come back and answer
your question here the question is what
in the world does this stuff mean what
are we doing here are we still finding
derivatives and what is it derivative me
well you have those words backwards
metal yeah kind of so do you mean the
slope of a point no not really some of
our curve at a point yes so if I say
here's Y squared minus X plus 1 equal 0
definitely an implicit equation can you
find the slopes of that curve at two
different points
find slopes at 2 negative 1 & 2 1 notice
something interesting that happens to
this because I have something that's not
actually a function it's implicit unplug
it in the same x-value I'm actually
getting out 2 y values you see that also
notice that I am implicit which means I
have to give you both x and y for you to
accomplish this problem because your
derivative is going to be in terms of
both x and y you see it so that's how
you use this now let's go ahead and end
this problem real quick this is a pretty
easy derivative if I derive both sides
I'm going to get DDX to the left side
equals DDX on the right side
start with each side what's the
derivative of 0 please good derivative
of Y squared perfect - how much 1 plus
come with a good derivative here is 2y
dy/dx through here is 1 derivative here
nothing can you solve that if you add 1
you're going to get 2y dy/dx equals 1 if
you divided by 2y
you get dy/dx equals 1 over 2y
what did you just solve for what is this
thing what is it that's M yeah that's it
that's look that's what that is so this
is a slope function now it says find the
slope at 2 negative 1 & 2 1 is there a
place to plug in the X not even a place
to plug it in I don't even care about
that plug in whatever values you can if
they had both X's and Y's you plug in
both X's and Y's bez only X's plug in X
is that's wise well you're wise
so what we're trying to find is a slope
add to negative 1
four to negative one I would have 1 over
2 times negative 1 4 to 1 and 1 over 2
times 1 this one gives me negative 1/2
this gives me positive 1/2 interestingly
here's what they've done we found the
slope of that thing at the two points so
two different slopes because we're going
to have two different outputs for that
same input interesting question how many
people are so we talked about so far
today ok I'm going to start you off next
time with quite a brain buster after
that we're done there's really one more
problem that we have to do so I want to
show you a couple things about that now
what's R on the next time alright so
today like I said we're going to
continue talking about implicit
differentiation that's what we're doing
we're still finding a derivative but
we're realizing that we can take this
implicitly so if we can't solve our
function for y like that which this is
actually equation not a function but if
we can't solve it for y that's called
implicit so this is definitely an
implicit equation where you don't find
the derivative which still means slope
of that thing so here's the question the
question is not just this but can you
with this equation finds the tangent
line at a certain point find the
equation of the tangent line
at the point two one by the way with
something like this could I just say at
X equals 3 or X equals 2 could I have
done that here now really if you can't
solve for y because even if I plug in
the the two you'd have a bear of a time
find out what the Y would be does that
make sense so really with these implicit
ones they're going to give you the full
point we're going to say that the
derivative as well so let's go ahead
let's find the equation of tangent line
at that point what do you need to do to
find the equation of a tangent line good
because for any line you need two things
one of the two things you need to find
the equation of the line slope and a
point we have a point we now just need a
slope a slope is a derivative so we're
going to go through this I want your
help with it but I'm going to do it on
the board for the sake of time just so
you see one more time the implicit stuff
I've already taught you how to do all
that so right now it's really just
practice for you and when one thing
you're going to see if it's very
important as far as our signs goes so
with the implicit differentiation says
you take a derivative DDX of both sides
and we treat Y as a function of X what
that means is Y is a function in terms
of X that we just don't happen to know
so if we're taking this with respect to
the variable X and Y is a function of X
we're going to get a dy/dx every time we
touch a Y with the derivative does that
make sense to you so we're going to be
getting that so in our case well here's
all the works shown for you we're going
to take a derivative of both sides
even though that's a zero I'm still
going to show the derivative just for
the sake of well-being correct and so
you don't forget that even if it's
constant or something you should have
take a derivative that so we're taking a
derivative of both sides of course a
derivative of zero is still zero so
we're good on that now let's look real
careful at these pieces over here is
this F our derivative take the four to
the fourth what's the derivative of 4x
to the fourth ok so let's do that so we
got that that derivative how about this
one is this one different for us that's
a project why is it a product rule sure
so this is a function in terms of X
right there that is an X now we said Y
is a function of X as well so every time
you get that you do need product rule or
if you have a fraction you would need
the quotient rule so you have to have
that so in our case right here we do
have a plus so we write a plus now
there's a couple things you can do with
that 8 you can treat it as 8x squared
that's one term or you can pull the 8
since that's a constant you can pull
that out front that really wouldn't
matter
now you're probably going to be
distributing that anyway so leave it in
there with the 8x squared that's fine
but you could you could pull it out DC
I'm talking about that's possible so
here I'll just treat it as 8x squared
that seems to work ok for me
so derivative the first times the second
plus the first times the derivative of
the second by the way note that even
when I'm taking a derivative of Y I
still have a DDX do you guys see that I
still have the DX
you can't go switching variables so if
you have if you start with the DDX it's
got to be that all the way through all
right so that's there no matter what
still okay so far so we've just done the
product rule you know what I'm going to
do since this piece was right here just
for you say something why do this and
next one just for the sake of being have
a cut and continued continuity in the
way I'm going to do things here I'm
gonna put a big bracket around that
that's saying this came from that one
term and that's a great thing to do the
reason why it's a great thing to do
you're going to see this in the next
piece if you do if you do this here it
really doesn't matter because that
positive isn't going to change any signs
you see I'm saying however what's going
to happen if I don't do that with this
one I'm going to really blow my problem
up okay that's going to explode my
problem because the sign is going to be
incorrect you're going to see that right
here let's look at the next term I have
a minus sign so I'll write of minus 25 x
squared Y that's still a product rule
product rule says maybe take the 25 with
the x squared d DX of x squared that's a
trivet of the first sorry
derivative of the first times a second
plus the first times the derivative of
the second
and once you see something now when this
said take the derivative of this piece
it means I'm going to subtract the
derivative of that piece the derivative
of this piece isn't just this thing is
it it's the whole thing so if you don't
have a big bracket right here that signs
it correct right if you do have a big
bracket right there that's saying that
what's going to happen is this negative
is going to change those signs in there
and that would be the correct way to do
it how people see that okay that's an
important thing that's actually why we
did this example just just so you can
see that otherwise you're going to be
off by a sign somewhere and then that's
a good place for plus plus let's do this
piece what's that piece okay I'll give
you a 16 y cube we do we need to do Y DX
on that every time you touch a Y with
the derivative you have to have a dy/dx
on account of what rule is that it did
that general power rule or in general
the chain rule that's right the chain
rule says that equals zero now that's a
long problem did you think you'd ever
get problem that long that's ridiculous
and it's like a whole paper right there
are you okay on what we've done so far
so we've done real easy one no problem
pluggable product rule dy/dx because
we're touching away with a derivative
and if we have zero yes no now we start
following the DDX this is why I made you
write the DDX on your problems because
it will tell you what to do next it says
this piece is done you're just going to
have the 16 X cubed plus now this piece
in here this just says do this DDX so
every one 8x squared what's the
derivative of that and then it's y
squared notice there's no dy/dx because
I don't need one here this is just the
derivative of X plus I'm getting rid of
my bracket because I know none of those
signs are going to change everything was
positive that's okay so this plus
doesn't change
that's where those brackets are
disappearing too then plus I have ax
squared tell me the derivative of Y
squared please when I'm taking a
derivative with respect to X what is
that good - Y but not just to Y right if
I do two if I just do to why you know
what happens if I do this and I forget a
dy/dx what's going to happen is you're
going to be adding and subtracting this
term over instead of dividing by that
term because you factored out a dy/dx do
you see that the difference there that's
going to really make that ending
fraction which we're probably going to
get look really weird so it's really
important to get that dy/dx because that
gives you another term that you're going
to factor dy/dx out of - okay let's see
about here the derivative of this this
guy and then I have a Y now when I've
already gotten rid of this bracket are
you guys who can do it a couple pieces
at once we get rid of this bracket
what's gonna happen to that so this
isn't math magic right I'm not Harry
Potter gone no more pleasure to swap off
a halt no and what's happening is I'm
getting rid of bracket by distributing
to these pieces notice the bracket ends
here that - won't affect this + but it
will affect this + so 50 XY plus our
sorry - 25 x squared + the derivative of
Y everybody very good then I'll have a +
16 Y cubed
dy/dx equals zero give a little quick
head not if you're okay on this so far
yeah okay oh my gosh Wow what now well
done yeah I see what you were no not
even close
well okay if it's done give me the slug
then because we want the equation
tangent line this is our goal right this
is not our goal get all the right hands
from one side brilliant yeah if we get
all our dy/dx terms on one side then we
can factor that if you get everything
else on the other side that's great
because ultimately we'll be able to
divide follow so let's get all of our
dy/dx is low we'll leave those on the
same side and then we're going to add or
subtract our terms depend on their sign
to eat them over so we will be
subtracting our 16x cubed our sixteen XY
squared will leave this one alone so we
want to leave this guy right here will
be getting rid of this one we want to
leave this one with the sign and leave
that one
by the way are you okay that I'm going
to make this 16 x squared Y you're out
with that too so notice that this is 16
x squared y dydx make a little bit nicer
looking so 16 x squared y dydx that's
that one so I got that minus 25 x
squared dy DX I got that one plus 16 y
cubed dy DX I got that one equals well
we have to add this and subtract these
two so you can write this in whatever
you want it really doesn't matter we're
going to have these three terms on the
other side the signs will look different
because we're adding or subtracting them
so if I add this over I get 50 X Y if I
subtract these two I'm going to get
minus 16 X cubed minus 16 XY squared
are you still okay so far okay great
dy DX's are all on one side everything
else is on the other side what now let's
factor so if i factor dy DX notice how
when we did this every term as it should
as a dy/dx in on the left hand side well
factor that out means divide everything
else I'm a big bracket
and now we have dy/dx times this
enormous factor how do you get rid of an
enormous factor being multiplied by some
dividing you can divide by this entire
expression right here that entire factor
is going to go under this expression on
the right hand side not mess that up so
dy/dx is this thing
all over that thing
okay quick show of hands how many people
were able to follow that pill okay with
it by the way did I do it right okay
okay
because I haven't had to do anything all
day so this is my first actual teaching
class on my brain silver rusty so
hopefully I did all that fight or maybe
I'm extra good today because I had to
waste my time on those stuff
oh you're getting extra your lesson
today I like that book okay so we have
this by the way you should try to
simplify this if you can factor and
simplify now you're seeing an X here but
you can factor out but there is not an X
comment all through those terms so it
really wouldn't help you in this case
also you look for any constants to
factor out you really don't have you
have a two here but not a two there so
there's nothing really much that we can
do do you see when talking about what
did we just find slope the equation for
the slope the formula for slope now we
said find the equation of the tangent
line at that point
do you have X's and Y's they give you X
and one look at this this is the slow
honey that's so it says plug in the
numbers and find the slope
and I have no idea what it's going to be
so okay do this correctly
things me nasty though looks horrible
I give up this way for you I get to 100
over 66 upgrade
I don't know what I got yet
I got briefed I don't have a calculator
shame on you did you without a
calculator
awesome awesome
no really nice
is that weird you're like I expect
negative 1257 over like a third well if
you put all this and you think this is a
hard question it's really not this is
the easy part this is a hard part let's
find the slope because once you I mean
if you talk about slope a slope has to
be a number right a slope s of line has
to be a constant number so plugging in
Bank a constant number now this is very
easy Y minus y1 equals M X minus x1 you
get Y minus 1 equals 3x minus 6 and the
1 is your answer Jaffee this doesn't
come about it all right oh by the way
don't plug this in the end not the right
way to go use that tip under slope does
it make sense to you so this is about
the hardest derivative I've ever really
seen for implicit unless I threw a
quotient rule that you have within that
some of them are harder but this is good
if you can do that you're fine okay do
you do that and one that involves a
quotient rule you're fine I think I did
give you one of those earlier in our
lessons so refresh your memory on that
should be the second derivative of
something I think we did that showed you
how to plug in that dy/dx set this ends
our implicit different differentiation
are you okay with the idea of implicit
differentiation
uh no loopy murder
I mean civil I told you before the
Senate
very very hard he's on his own so am I
gonna give you a problem that takes you
45 minutes doing a test maybe they won't
do that to you
unless I just give you a test that has
one problem on this I feel a great
advise and say this is your final
congratulations
haha is it that you drink my coffee but
I probably would wouldn't you probably
won't do that but you want to roll it
out either probably greater than 50%
chance that I won't do that
anyway ji-sung idea of implicit
differentiation now what's kind of cool
is that we can use this idea of implicit
differentiation to take a derivative of
whatever we want
with respect to whatever variable we
want and that's a weird idea but here's
how we can use it I'm going to introduce
the idea of something called a related
rate these are all word problems I know
some of you have a just it's the bane of
your existence or word problems if you
think through I'll give you some steps
on how to do these and they shouldn't I
can't of course I can't give you an
example of every type of word properties
lots of word problems but what I will do
is give you a technique on how to do
these when they write problems are you
ready for it so if I'm at 2.8