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Implicit differentiation And the Product Rule
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Feb 19, 2025
In mathematics, implicit differentiation is the process of finding the derivative of a function that is not given in an explicit form. In other words, it is the process of differentiating a function that is not given as y = f(x). This may seem like a daunting task, but luckily there is a rule that we can use to make the process much easier. This rule is called the product rule. In this video, we will learn how to use the product rule when performing an implicit differentiation.
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0:00
all right so let's work on another
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problem with the product rule
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and the implicit colorful all right
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so we're gonna find the derivative of
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x cubed
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y cubed minus 3y squared
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equals to two
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so what do we do in this case
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we have to use the protocol okay so
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let's try to split this again just like
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we did the first one right so
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one let's start with the easy one so if
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I have
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negative 3y squared what's the
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derivative of that
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negative 6y and
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oh um DX I mean d y over DX d y over DX
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implicit right and now for this one
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we're going to use the protocol so we
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have what
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um two x cubed
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times y cubed so what will be the
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derivative here
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so I'm going to start with x cubed right
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F Prime G Plus G Prime F so what's the
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derivative of x cubed
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3x square right times y cubed Plus
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the derivative of y cubed is coming 3y
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squared times what d y over DX and then
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times x cubed times x cubed right and
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now now that we find it separately we
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can put them all together so we start
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with this one
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so that's going to be three
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x squared
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y cubed
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Plus
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3y squared
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d y or DX
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times x cubed
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minus
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6y
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d y over DX and what's the derivative of
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2
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0 0 right and now you can just collect
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these and put them together so we put d
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y over DX as a factor we have d y with
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DX
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times
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3y squared
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minus
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2y squared x cubed minus 6y
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equals to
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zero okay
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and
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that's it pretty much
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okay
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so in this case d y over DX is just
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going to give you zero because if you
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divide everything by zero just don't
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give you zero right
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so we can stop here so by DX
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if you shift this you divided by it's
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going to be gone
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right so that's just going to give you
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zero
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so that's the protocol and uh
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the implicit on differentiation
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wait what happens to 3x squared y cubed
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3x square y cube is right here oh you're
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right it's not going to be zero I'm okay
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I got you guys got it
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yeah I was kind of thinking yeah
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something happened that means you guys
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are paying attention because 3x squared
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y cubed equals to zero we're going to
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shift this to the other side so we got d
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y over DX
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3y squared x cubed
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minus 6y equals negative 3x squared
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y cubed and then we divide this
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all right so this cancels out so we have
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d y
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would be x equals negative three x
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squared y cubed over 3y squared x cubed
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minus 6y
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so that is the product rule
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well if you were to simplify that can
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you just
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you can take out the Y's yeah you can
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simplify that if you put y as a factor
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actually you can simplify it not only
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just y 3y as well right if I put three y
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as a factor I'm going to be left with
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what negative x squared y squared right
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over
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3y that gives me y x squared x cubed
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minus
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um
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two right so this cancels out so you're
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going to be left with negative x squared
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y squared over y x cubed
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minus 2. I'm more interested in like
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understanding the concept but yes you
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must always
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simplify you don't leave stuff
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unsemplified
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especially if you're doing calculus
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so now I have an interesting function
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for you let's do some trigonometric
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functions
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it's getting more interesting right
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yeah
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I'm trying to learn trigonometry
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oh yeah
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actually there's another one that I want
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to do before we jump onto the trig
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functions that's the trick functions
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tomorrow I wanna I want you guys to sit
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on this one today
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so
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are you all done copying this all right
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so we have x squared y
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plus y squared x equals I think
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is negative two
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so find uh the derivative so I'll give
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you some time to to come up with the
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answer
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we have x square y
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plus y squared x equals to negative 2.
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so basically we're gonna do the same
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thing right we're gonna split this okay
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so what's the derivative of just x
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square y using the product rule
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so what would it be
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yep
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so first the derivative of this right 2x
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times what Y and then Plus
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times x squared excellent all right and
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then
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second
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the
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so what's the derivative
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um 2i times x
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oh wait no uh d y over DX all right
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times n wait yeah that's good yes
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one times y squared that's it good got
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it so that's going to give us the first
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one so we have 2xy
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Plus
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d y
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over DX
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x squared and then plus this one which
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is 2y
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x d y over DX
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and then plus y squared
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and the derivative of negative 2 is
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obviously zero right right would there
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be a
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oh never mind okay and now we can
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collect the d y by D X is put it as a
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factor
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so if you have d y over DX
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is going to be times x squared plus 2y X
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right and then plus 2xy plus y squared
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equals to zero we're going to shift this
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to the right side
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so we have the Y for the X
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times x squared plus 2y X
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equals to negative 2 X Y minus y squared
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and then the last step we just to divide
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both rows
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thank you
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again
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now in this case you will not be able to
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factor out any excess of you can find
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out the X1 then nothing's going to
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cancel out so you have to leave it this
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way okay that's nothing else you can do
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at this point wait what once you get
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here yeah okay you can stuck here
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because X does not come with a Y Square
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so there's not much you can do you have
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to stop at this point okay so that's the
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protocol
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and are implicit now we're going to talk
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about a quotient rule I think
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unfortunately we should do that tomorrow
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and then three functions tomorrow so I
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want to sit on this a little bit okay
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any questions
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good
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