### Video Transcript

The equation of a curve is given as
two π¦ squared minus π₯ squared is equal to negative 16π¦ minus 36. Part one) Find dπ¦ by dπ₯ in terms
of π₯ and π¦. Part two) Write an equation for the
line tangent to the curve at the point six, zero. Part three) Write an equation of
each vertical tangent to the curve. Part four) Evaluate d two π¦ by dπ₯
squared at the point three root six, one.

For part one of this question,
weβre asked to find dπ¦ by dπ₯ in terms of π₯ and π¦. Our equation is two π¦ squared
minus π₯ squared is equal to negative 16π¦ minus 36. We can find dπ¦ by dπ₯ by using
implicit differentiation. This means we simply differentiate
the whole equation with respect to π₯. We can do this term by term for
each term in the equation. Weβll start with the term two π¦
squared. We need to find the differential of
two π¦ squared with respect to π₯.

In order to do this, we require the
chain rule. We can recall that the chain rule
tells us that dπ by dπ₯ is equal to dπ by dπ¦ multiplied by dπ¦ by dπ₯. If we let two π¦ squared here be
equal to π, then we can say that our differential here is equal to d by dπ¦ of two
π¦ squared multiplied by dπ¦ by dπ₯. We can find d by dπ¦ of two π¦
squared by using the power rule for differentiation. We multiply by the power and
decrease the power by one. We obtain that the differential of
our first term is four π¦ multiplied by dπ¦ by dπ₯.

The second term in our equation is
negative π₯ squared. And we can differentiate this with
respect to π₯ by using the power rule for differentiation. We obtain negative two π₯. Now moving to the right-hand side
of the equation, our next term is negative 16π¦. When we differentiate π¦ with
respect to π₯, we simply get dπ¦ by dπ₯. And we mustnβt forget to multiply
it by our constant. This gives us negative 16 dπ¦ by
dπ₯. Our final term negative 36 is a
constant. And when we differentiate any
constant, weβll get zero. Therefore, we have finished
differentiating our equation with respect to π₯.

Now, we simply need to make dπ¦ by
dπ₯ the subject. First, we divide the whole equation
by two. Then, we move all the terms
containing dπ¦ by dπ₯ to one side of the equation. Then, we factor out dπ¦ by dπ₯. And finally, we divide to obtain
our solution to part one that dπ¦ by dπ₯ is equal to π₯ over two π¦ plus eight.

For part two, weβre required to
write an equation of the line tangent to the curve at the point six, zero. Now since the tangent is going to
be a straight line, we know that it will be of the form π¦ is equal to ππ₯ plus π,
where π is the slope of the tangent. And we can find the slope of the
tangent using the equation we found in part i since dπ¦ by dπ₯ can give the gradient
of the curve at any point. Now we need to find the slope of
the curve at the point six, zero.

We can simply substitute π₯ equals
six and π¦ equals zero into our equation for dπ¦ by dπ₯. In doing this, we obtain that dπ¦
by dπ₯ which is also the slope of our tangent is equal to three-fourths. Therefore, we can substitute π
equals three-fourths into our equation for our tangent. Now all that remains is to find the
value of b. We can do this by substituting π₯
equals six and π¦ equals zero into the equation for our tangent. Rearranging this, we find that π
is equal to negative nine over two. Now, I put this back into the
equation of our tangent to obtain an equation of π¦ is equal to three-fourths π₯
minus nine over two. This is the equation of the tangent
to the curve at the point six, zero.

Part three) Write an equation of
each vertical tangent to the curve.

Now, a vertical line has an
infinite slope. And so therefore, our vertical
tangent will occur when the denominator of dπ¦ by dπ₯ is equal to zero since dπ¦ by
dπ₯ is the equation for the slope of our curve. The denominator of dπ¦ by dπ₯ is
two π¦ plus eight. So we can set that equal to
zero. Now, we rearrange this. And we find that π¦ is equal to
negative four. Now, this is an equation of a
line. However, this is a horizontal
line. And we have been asked to find the
vertical tangents.

In order to find the vertical
tangents, we can substitute this value of π¦ is equal to negative four back into the
original equation. We obtain this which we can
simplify and then rearrange to obtain that π₯ squared is equal to four. Now, we will take the square root
here. But we must be careful to remember
that we would obtain both positive and negative solutions. We get π₯ is equal to negative two
and π₯ is equal to two. Now, these are equations of
vertical lines. Therefore, π₯ equals negative two
and π₯ equals two are the equations to the vertical tangents of our curve.

Part four) Evaluate d two π¦ by dπ₯
squared at the point three root six, one.

Now, d two π¦ by dπ₯ squared is the
second differential of π¦ with respect to π₯ or the differential of dπ¦ by dπ₯ with
respect to π₯. Now, weβve already found dπ¦ by dπ₯
in part one of the question. And so, in order to find d two π¦
by dπ₯ squared, we simply need to differentiate π₯ over two π¦ plus eight with
respect to π₯. Here, weβre differentiating a
quotient. And so, we can use the quotient
rule. The quotient rule tells us that d
by dπ₯ of π’ over π£ is equal to π£ dπ’ by dπ₯ minus π’ dπ£ by dπ₯ all over π£
squared.

If we compare this to our equation,
we can say that π₯ is equal to π’ and two π¦ plus eight is equal to π£. Next, we need to find dπ’ by dπ₯
and dπ£ by dπ₯. To find dπ’ by dπ₯, we
differentiate π₯ with respect to π₯. This simply gives us one. Now, to find dπ£ by dπ₯ when you
differentiate two π¦ plus eight with respect to π₯, we obtain that dπ£ by dπ₯ is
equal to two dπ¦ by dπ₯. Now, we can use the quotient rule
and substitute in these values. We obtain that d two π¦ by dπ₯
squared is equal to two π¦ plus eight multiplied by one minus π₯ multiplied by two
dπ¦ by dπ₯ all over two π¦ plus eight squared.

Now, weβre asked to evaluate d two
π¦ by dπ₯ squared at the point three root six, one. So our next step here would be to
substitute in π₯ equals three root six and π¦ equals one. However, we would not know the
value of dπ¦ by dπ₯. So letβs start by finding the value
of dπ¦ by dπ₯ at the point three root six, one. We substitute these values into the
equation we found in part one. And then, we can simplify this to
obtain that at the point three root six, one, dπ¦ by dπ₯ is equal to three root six
over 10.

Now, we can evaluate d two π¦ by
dπ₯ squared at the point three root six, one by substituting in the values for π₯,
π¦, and dπ¦ by dπ₯. And next, we can simplify this to
obtain that at the point three root six, one, d two π¦ by dπ₯ squared is equal to
negative one over 125. This completes the final part of
the question.

The solutions we found are as
follows. Part one) dπ¦ by dπ₯ is equal to π₯
over two π¦ plus eight. Part two) π¦ is equal to
three-fourths π₯ minus nine over two. Part three) π₯ is equal to negative
two and π₯ is equal to two. And finally, for part four, we
evaluated d two π¦ by dπ₯ squared to be equal to negative one over 125.