### Video Transcript

Given that π₯ squared plus nine
equals negative two π₯π¦, find π₯ times d two π¦ by dπ₯ squared plus two dπ¦ by
dπ₯.

It should be quite clear that to
work this out, weβre going to need to find dπ¦ by dπ₯, thatβs the first derivative
of π¦ with respect to π₯, and d two π¦ by dπ₯ squared. Thatβs the second derivative of π¦
with respect to π₯. If, however, we go over to our
equation, we notice itβs not so easy to differentiate π¦ with respect to π₯. And this is because this function
is defined implicitly. Weβre used to differentiating
functions, so π¦ is given explicitly as a function of π₯. Luckily, we can use a special
version of the chain rule to differentiate a function in π¦ with respect to π₯. When we do, we differentiate that
function with respect to π¦ and then multiply that by dπ¦ by dπ₯.

So letβs take the equation π₯
squared plus nine equals negative two π₯π¦. Weβll differentiate both sides of
this equation with respect to π₯. To do so, we recall how we actually
differentiate a power term. For a term of the form ππ₯ to the
πth power, where π is a real constant, we multiply the entire term by π and then
reduce that power or that exponent by one. This means that the derivative of
π₯ squared is two π₯ to the power of one or just two π₯. It also means that the derivative
of a constant is zero. Remember, nine can be thought of as
nine π₯ to the power of zero. And when we multiply all of that by
the exponent zero, we get zero.

But what about differentiating
negative two π₯π¦ with respect to π₯? Itβs actually the product of two
differentiable functions. So weβre going to use the product
rule combined with implicit differentiation. This says that the derivative of π’
times π£ is π’ times dπ£ by dπ₯ plus π£ times dπ’ by dπ₯. In our function, weβre going to let
π’ be equal to negative two π₯ and π£ be equal to π¦. Then, dπ’ by dπ₯, the first
derivative of negative two π₯ with respect to π₯, is simply negative two.

But what about dπ£ by dπ₯? Weβre going to use implicit
differentiation. We know we need to differentiate
our function with respect to π¦. Well, the derivative of π¦ with
respect to π¦ is just one. We then multiply that by dπ¦ by
dπ₯. So we can see that dπ£ by dπ₯ is
one dπ¦ by dπ₯ or just dπ¦ by dπ₯. Then, the derivative of negative
two π₯π¦ is π’ times dπ£ by dπ₯. Thatβs negative two π₯ dπ¦ by dπ₯
plus π£ times dπ’ by dπ₯, which gives us negative two π¦. And so weβve differentiated both
sides of our equation with respect to π₯. We get two π₯ equals negative two
π₯ dπ¦ by dπ₯ minus two π¦.

Letβs add two π¦ to both sides of
our equation, and we get two π₯ plus two π¦ equals negative two π₯ dπ¦ by dπ₯. Next, we find an expression for dπ¦
by dπ₯ by dividing through by negative two π₯. We see dπ¦ by dπ₯ is two π₯ plus
two π¦ over negative two π₯. Now, in fact, we can simplify this
to get negative one minus π¦ over π₯. So we have an expression for dπ¦ by
dπ₯. Letβs repeat this process and find
an expression for d two π¦ by dπ₯ squared. Letβs differentiate our expression
for dπ¦ by dπ₯ term by term. Well, the derivative of negative
one is actually zero. And this means d two π¦ by dπ₯
squared is simply the derivative of negative π¦ over π₯ with respect to π₯. And, of course, we could, if we
choose, write this as negative the derivative of π¦ over π₯ with respect to π₯. But how are we going to
differentiate π¦ over π₯?

We know that π¦ and π₯ are
differentiable functions. So weβre going to use the quotient
rule. This says that the derivative of π’
divided by π£ is π£ dπ’ by dπ₯ minus π’ dπ£ by dπ₯ all over π£ squared. Now if we compare this to our
function, we can see that weβll let π’ be equal to π¦ and π£ be equal to π₯. dπ£ by
dπ₯ is quite straightforward. Itβs the derivative of π₯ with
respect to π₯, which is just one. But once again, weβll need to use
implicit differentiation to find dπ’ by dπ₯. Itβs the derivative of π¦ with
respect to π¦, which is one, times dπ¦ by dπ₯. So in fact, itβs simply dπ¦ by
dπ₯.

The quotient rule tells us that the
derivative of π¦ over π₯ with respect to π₯ is π₯ times dπ¦ by dπ₯, thatβs π£ times
dπ’ by dπ₯, minus π’ times dπ£ by dπ₯, which is π¦ times one. And, of course, thatβs all over π₯
squared. We simplify this a little. And then we see that the derivative
of π¦ over π₯ is π₯ dπ¦ by dπ₯ minus π¦ over π₯ squared. So d two π¦ by dπ₯ squared is the
negative of this. We distribute our parentheses by
multiplying through by negative one. And we see that d two π¦ by dπ₯
squared is π¦ minus π₯ dπ¦ by dπ₯ over π₯ squared.

But, of course, we actually have an
expression for dπ¦ by dπ₯. Itβs negative one minus π¦ over
π₯. And so we replace dπ¦ by dπ₯ with
this. And we see that d two π¦ by dπ₯
squared is π¦ minus π₯ times negative one minus π¦ over π₯ all over π₯ squared. And then we distribute the
parentheses. Negative π₯ times negative one is
π₯. And negative π₯ times negative π¦
over π₯, well the π₯βs cancel, and we get positive π¦. And so d two π¦ by dπ₯ squared is
two π¦ plus π₯ all over π₯ squared. We now have expressions for the
first and second derivative of π¦ with respect to π₯. So letβs substitute them into this
expression.

We get π₯ times two π¦ plus π₯ over
π₯ squared plus two times negative one minus π¦ over π₯. We can see that we can cancel a
factor of π₯. And the first part of this becomes
two π¦ plus π₯ over π₯. It also then makes sense to
distribute our parentheses by multiplying each term by two. And so we get negative two minus
two π¦ over π₯. Weβre actually gonna split our
first fraction up by dividing each bit by π₯. We get two π¦ over π₯ plus π₯ over
π₯, which is just one. And now we see we have two π¦ over
π₯ minus two π¦ over π₯, which is zero. And so this actually simplifies to
one minus two, which is negative one. And so π₯ times d two π¦ by dπ₯
squared plus two times dπ¦ by dπ₯ is negative one.

Now, itβs worth noting that we
could have originally divided both sides of our equation by negative two π₯. We then couldβve used the quotient
rule twice to find dπ¦ by dπ₯ and d two π¦ by dπ₯ squared. By dividing by negative two π₯, we
end up with an equation thatβs explicitly in terms of π₯.