Here we're given an implicit equation,
and asked to find the dy dx,
then determine the equation the tangent line
at the point one comma one.
We have an implicit equation here
because notice how it's not solved for y in terms of x.
So the first type is a different shape both sides
of the equation with the respect of x.
And now when we differentiate with respect to x,
if the term only contains x,
we differentiate as we normally would,
but if it contains y, we have to apply the chain rule,
and we have an extra factor of dy dx.
Looking at the left side, we first want
to find the derivative of negative 54 x to the ninth,
so we multiply by nine;
nine times negative 54 is negative 486.
We subtract one from the exponents,
so that'd be x to the eighth,
plus now we want to find the derivative
of eight x to the 36th times y,
because this term contains both x and y,
we need to think of this as a part
of two functions and apply the product rule.
For review, the product rule is given here below.
So let's let eight x to the 36th be the first function, f,
and we'll let y be the second function, g.
And now applying the product rule,
we have the first function as it is,
so plus eight x to the 36th times
the derivative of y with respect to x,
because this is a y term,
we have to apply the chain rule,
and therefore we'll have an extra factor of dy dx.
So first we find the derivative of y with the respect to y,
which should be one, and then times a factor of dy dx,
which would just give us dy dx.
Plus, the second function as it is, which is y,
times the derivative of the first function,
the derivative of eight x to the 36th,
again this only contains x, so we multiply it by 36.
36 times eight is equal to 288,
so we have times 288
x to the 36th minus one is 35,
plus the derivative of y to the tenth,
again this is a y term, so we have to apply the chain rule,
so we first find the derivative with the respect to the y,
then multiply by dy dx,
so we'd have 10 y to the ninth times dy dx
equals a derivative of negative 45, which is zero.
Now we want to solve this equation for dy dx.
Notice how these two terms contain a factor of dy dx,
and these two terms don't.
So let's move the two terms that don't have a factor
of dy dx to the right side,
so we have eight x to the 36th dy dx
plus 10 y to the ninth times dy dx,
and this would be equal to
we'll add 486 x to the eighth to both sides,
so we have a positive 486 x to the eighth on the right,
and now also drag this term to both sides,
so we'd have minus 288
x to the 35th y.
And now we'll factor out the new dy dx,
which will leave us with eight x
to the 36th plus 10 y to the ninth,
equals the right side remains the same,
and now divide both sides by the quantity eight x
to the 36th plus 10 y to the ninth.
Notice how this simplifies to one, and therefore dy dx
is equal to 486 x to the eighth
minus 288 x to the 35th y
divided by the quantity eight x to
the 36th plus 10 y to the ninth.
While this derivative is correct, it does simplify.
Notice how all these terms
do contain a common factor of two.
Let's begin by factoring out two
from the numerator and denominator.
Factor out two from the numerator would have
two times the quantity
243 x to the eighth
minus 144 x to the 35th y
divided by factoring out a two
from the denominator leaves us with
two times the quantity four
x to the 36th plus five y to the ninth.
So two over two simplifies to one.
This isn't going to simplify any further,
though sometimes we may see the numerator factored,
so we have dy dx
equals 243 x to the eighth
minus 144 x to the 35th y
divided by four x to the 36th
plus five y to the ninth,
but we may also see it expressed
where the numerator is factored.
The greatest common factor of the numerator would be
nine x to the eighth,
which would leave us with 27
minus 16 x to the 27th y,
and the numerator would stay the same.
We have four x to the 36th
plus five y to the ninth.
So I just want to point out that it would not be wrong
to leave the derivative in this form here,
but if you're looking in the back of a book,
it would probably express a derivative
in either of these two forms here.
Now that we have dy dx, we can find the slope
of the tangent line at the point one comma one
by evaluating this derivative at this point,
so dy dx,
at the point one comma one, is equal to,
well if we substitute one for x and one for y,
let's use this form of the derivative here,
if both x and y are one,
the numerator is just going to be 243 minus 144
divided by our denominator is going to be four plus five.
This is going to give us positive 99
divided by nine, which equals 11.
So now we want to find the equation of the line
that has a slope of 11,
and passes through the point one comma one.
Let's do this on the next slide.
We are asked to express the equation
of the tangent line in sloped intercept form,
which would be the form y equals mx plus b.
So to find the linear equation, we can use this form,
or if we wanted to, we could begin with point slope form,
which is y minus y equals m
times the quantity x minus x of one.
Either form works.
Let's go and just use slope intercept form for this example.
So we know the slope was 11,
so we know the equation must be in the form
of y equals 11 x plus b,
and then because the point one comma one is on this line,
it must satisfy this equation,
so now if we substitute one for x and one for y,
we can determine b, the vertical intercept,
or y intercept.
We would have one equals 11 times one plus b,
or one equals 11 plus b,
subtracting 11 on both sides,
we have b equals negative 10.
So the equation where tangent line would be y equals,
we know our slope is 11, so 11 x,
and now we know the vertical intercept is negative 10,
so we have minus 10.
So let's just verify we get the same thing
if we use point slope form.
In point slope form, we'd have the equation y minus
y one would be one equals m, which we know is 11
times the quantity x minus x of one.
X of one is one.
So we have y minus one equals,
distributing, we have 11 x minus 11,
adding one to both sides, we get the same equation,
y equals 11 x minus 10.
I hope you found this helpful.