hi everyone today we're going to talk about how to use implicit differentiation to find
the equation of the tangent line at a given point to complete this problem we'll use implicit
differentiation to find the first derivative then we'll plug the given point into the derivative
to find the slope of the tangent line finally we'll plug the slop and given point into the
point slope formula for the equation of the line to find the equation of the tangent line
let's take a look in this particular problem we've been asked to use implicit differentiation
to find an equation of the tangent line to the curve at the given point and as you can
see we have a function that is all tangled up in terms of x and y we can't easily separate
x and y so we really need to use implicit differentiation to find the derivative of
this function remember that the equation of the tangent line can be expressed in point
slope form where we have the equation as y minus y sub one is equal to m times x times
x sub one remember that we're looking for when we have the equation of the tangent line
in this form we're looking for three values we have y sub one which we can get from the
point that we've been given and that's already we know is going to be one half we have x
sub one which again comes from the point we've been given so we already know that that will
be zero the only thing we really need to find here is m and m is the slope of the tangent
line at the given point the only way to find the slope is by taking the derivative of the
equation and then plugging the point into the derivative so we're going to go at this
problem in a couple steps the first thing we're going to do is use implicit differentiation
to find the derivative of this equation then our second step will be to plug in the point
zero one half to our derivative and then that will give us m so once we have m we can plug
in x sub y sub one and m to our equation of the tangent line so let's go ahead and find
the derivative of this equation remember that when we're talking about implicit differentiation
we're talking about taking the derivative of this equation with respect to x so that's
important because what it means is that when we deal with terms that have only x involved
we don't have to worry about the derivative taking the derivative for example of x squared
with respect to x is the same kind of derivative that we're used to the derivative is two x
no problem but when we take the derivative of something like y square we have to use
chain rule and multiply by y prime because we have to treat y as a function so here's
what that looks like we'll take the derivative of one term at a time and the derivative of
x squared will simply be two x the derivative of y squared however will be two y just as
if y were the x variable but because we're taking the derivative with respect to x and
we have a y variable involved here we have to multiply our result here by y prime so
the easiest way to remember this is just that whenever you take the derivative of something
that involves y you have to multiply by y prime or the derivative y so let's see what
that looks like on our right hand side so we have equals notice that we have a quantity
here that's squared that means that we have to use chain rule and to use chain rule we'll
bring this two all the way out in front here and then we'll subtract one from the exponent
so the derivative of this using chain rule will be two times what's inside the square
is part of the outside function the two x squared plus two y squared minus x is the
inside function so we take the derivative of the outside leaving the inside completely
untouched we don't touch it so we get two x squared plus two y squared minus x and that
exponent goes away because we get two minus one which is just one so we took the derivative
of the outside function leaving the inside function alone but now according to chain
rule we have to multiply by the derivative of the inside function which means that we
have to take the derivative of what's inside now two x squared plus two y squared minus
x so the derivative of that term by term and again we're multiplying here the derivative
of two x squared will be four x because it's just x nothing fancy nothing fancy involved
with that one but the derivative of two y squared will be the same thing for y except
that we're taking the derivative of something that involves y which means we have to multiply
by y prime and it's as simple as that the derivative of negative x is just negative
one again since y is not involved we don't have to do anything special there and we just
get negative one so that's the derivative of the right hand side and now it's just a
matter of simplifying what we'll want to do eventually is solve for y prime so we're going
to need to get all of our terms that involve y prime on the left and everything that doesn't
involve y prime on the right so let's first go ahead and deal with this right hand side
here and multiply everything out so we get two x plus two y times y prime on the left
and then over here let's go ahead and distribute the two across this first term so we'll get
four x squared plus four y squared y squared minus two x and we'll be multiplying that
whole thing by four x plus four y y prime minus one now again we'll multiply those two
terms together and we'll just be foiling here so we'll take the first term and multiply
it by the first term then the first times the second and the first times the third then
the second times the first second and third etcetera so four x squared times four x will
give us sixteen x cubed four x squared times four y y prime will give us sixteen x squared
y y prime four x squared times a negative one will give us a negative four x squared
now we'll move onto our second term here four y squared so four y squared times four x gives
us sixteen x y squared four y squared times four y y prime gives us sixteen y cubed y
prime and then four y squared times a negative one gives us a negative four y squared now
we have negative two x here so negative two x times four x gives us a negative eight x
squared negative two x times four y y prime gives us negative eight x y y prime and then
negative two x times a negative one gives us a positive two x so now we can do some
quick simplification here notice we have a two x on the left hand side and a two x on
the right hand side they're both positive they're the same term so they'll cancel with
one another now we want to combine terms as much as we can and we also want to separate
the terms that involve y prime from the terms that don't so we have a two y y prime over
here on the left so that takes care of that one and I'll underline them as I go then I
want to move this sixteen x squared y y prime to the left hand side because it involves
y prime so I'll subtract it from both sides to get negative sixteen x squared y y prime
I'm looking for my next y prime here and I see that I have sixteen y cubed y prime so
I'll subtract that from both sides get sixteen y cubed y prime and then I have this term
here I have a negative eight x y y prime so I want to add it to both sides and I'll get
plus eight x y y prime those are all the terms involving y prime so that's everything that's
going to move to the left everything else will be left on the right hand side so I'll
have equals sixteen x cubed which takes care of this term then I have a minus four x squared
and a minus eight x squared so these two will combine with each other to get negative twelve
x squared then I have a plus sixteen x y squared so plus sixteen x y squared and finally a
minus four y squared okay so now what I want to do is factor out my y prime or my first
derivative of y so I'm going to pull that out and I have that multiplied by two y minus
sixteen x squared y minus sixteen y cubed plus eight x y so notice I just factored that
out and that's still equal to this right hand side
now I want to go ahead and do is divide both sides by this whole quantity here two y minus
sixteen x squared y etc to get y prime on its own so I'll get y prime is equal to the
right hand side sixteen x cubed minus twelve x squared plus sixteen x y squared minus four
y squared all divided by what I had over here on the left hand side
and now that I have it solved for y prime I want to go ahead and plug in my point zero
one half I'll plug in zero for x and one half for y and see what I get for y prime so I'll
get y prime equals sixteen times zero cubed minus twelve times zero squared plus sixteen
times zero times one half squared minus four times one half squared and then in my denominator
when I plug in I'll get two times one half minus sixteen times zero squared times one
half minus sixteen times one half cubed plus eight times zero times one half and you'll
notice I can go ahead and cancel everything with a zero in it so all these terms here
will cancel and I'll just be left with three terms I'll need to focus one so I can see
that I'll have y prime is equal to on the top here negative four times one half squared
we'll be negative four times one fourth and in my denominator on the bottom I'll have
two times one half which will just be one and I'll have sixteen times one half cubed
one half cubed will be one eighth so I'll have sixteen negative sixteen times one eighth
when I simplify this in the numerator my fours will cancel and I'll end up with negative
one in my denominator I'll have one minus sixteen times one eighth the eight will cancel
and I'll be left with sixteen over eight which is just two so I'll get one minus two of course
this will simplify to negative one over negative one which is just one so what I found here
is m by finding the derivative and plugging in the point zero one half I found the slope
of the tangent line at that point so m equals one which means I can return to my tangent
line equation and plug in the three values that I need so y minus y sub one which I know
to be one half from the point I was given I know that m is one and I know that x sub
one is zero again from the point we were given so now I just simplify this as much as possible
y minus one half is equal to just x on this side because I get x and then I get zero x
plus zero or x minus zero so I just get x and then I can go ahead and add one half to
both sides to get this into slope intercept form and I get y equals x plus one half and
this equation here is the equation of the tangent line to this curve our original equation
at the point zero one half so I you found that video helpful if you did like this video
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