Video Transcript
Given that 𝑥 squared plus nine
equals negative two 𝑥𝑦, find 𝑥 times d two 𝑦 by d𝑥 squared plus two d𝑦 by
d𝑥.
It should be quite clear that to
work this out, we’re going to need to find d𝑦 by d𝑥, that’s the first derivative
of 𝑦 with respect to 𝑥, and d two 𝑦 by d𝑥 squared. That’s the second derivative of 𝑦
with respect to 𝑥. If, however, we go over to our
equation, we notice it’s not so easy to differentiate 𝑦 with respect to 𝑥. And this is because this function
is defined implicitly. We’re used to differentiating
functions, so 𝑦 is given explicitly as a function of 𝑥. Luckily, we can use a special
version of the chain rule to differentiate a function in 𝑦 with respect to 𝑥. When we do, we differentiate that
function with respect to 𝑦 and then multiply that by d𝑦 by d𝑥.
So let’s take the equation 𝑥
squared plus nine equals negative two 𝑥𝑦. We’ll differentiate both sides of
this equation with respect to 𝑥. To do so, we recall how we actually
differentiate a power term. For a term of the form 𝑎𝑥 to the
𝑛th power, where 𝑛 is a real constant, we multiply the entire term by 𝑛 and then
reduce that power or that exponent by one. This means that the derivative of
𝑥 squared is two 𝑥 to the power of one or just two 𝑥. It also means that the derivative
of a constant is zero. Remember, nine can be thought of as
nine 𝑥 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 𝑥𝑦 with respect to 𝑥? It’s actually the product of two
differentiable functions. So we’re going to use the product
rule combined with implicit differentiation. This says that the derivative of 𝑢
times 𝑣 is 𝑢 times d𝑣 by d𝑥 plus 𝑣 times d𝑢 by d𝑥. In our function, we’re going to let
𝑢 be equal to negative two 𝑥 and 𝑣 be equal to 𝑦. Then, d𝑢 by d𝑥, the first
derivative of negative two 𝑥 with respect to 𝑥, is simply negative two.
But what about d𝑣 by d𝑥? We’re going to use implicit
differentiation. We know we need to differentiate
our function with respect to 𝑦. Well, the derivative of 𝑦 with
respect to 𝑦 is just one. We then multiply that by d𝑦 by
d𝑥. So we can see that d𝑣 by d𝑥 is
one d𝑦 by d𝑥 or just d𝑦 by d𝑥. Then, the derivative of negative
two 𝑥𝑦 is 𝑢 times d𝑣 by d𝑥. That’s negative two 𝑥 d𝑦 by d𝑥
plus 𝑣 times d𝑢 by d𝑥, which gives us negative two 𝑦. And so we’ve differentiated both
sides of our equation with respect to 𝑥. We get two 𝑥 equals negative two
𝑥 d𝑦 by d𝑥 minus two 𝑦.
Let’s add two 𝑦 to both sides of
our equation, and we get two 𝑥 plus two 𝑦 equals negative two 𝑥 d𝑦 by d𝑥. Next, we find an expression for d𝑦
by d𝑥 by dividing through by negative two 𝑥. We see d𝑦 by d𝑥 is two 𝑥 plus
two 𝑦 over negative two 𝑥. Now, in fact, we can simplify this
to get negative one minus 𝑦 over 𝑥. So we have an expression for d𝑦 by
d𝑥. Let’s repeat this process and find
an expression for d two 𝑦 by d𝑥 squared. Let’s differentiate our expression
for d𝑦 by d𝑥 term by term. Well, the derivative of negative
one is actually zero. And this means d two 𝑦 by d𝑥
squared is simply the derivative of negative 𝑦 over 𝑥 with respect to 𝑥. And, of course, we could, if we
choose, write this as negative the derivative of 𝑦 over 𝑥 with respect to 𝑥. But how are we going to
differentiate 𝑦 over 𝑥?
We know that 𝑦 and 𝑥 are
differentiable functions. So we’re going to use the quotient
rule. This says that the derivative of 𝑢
divided by 𝑣 is 𝑣 d𝑢 by d𝑥 minus 𝑢 d𝑣 by d𝑥 all over 𝑣 squared. Now if we compare this to our
function, we can see that we’ll let 𝑢 be equal to 𝑦 and 𝑣 be equal to 𝑥. d𝑣 by
d𝑥 is quite straightforward. It’s the derivative of 𝑥 with
respect to 𝑥, which is just one. But once again, we’ll need to use
implicit differentiation to find d𝑢 by d𝑥. It’s the derivative of 𝑦 with
respect to 𝑦, which is one, times d𝑦 by d𝑥. So in fact, it’s simply d𝑦 by
d𝑥.
The quotient rule tells us that the
derivative of 𝑦 over 𝑥 with respect to 𝑥 is 𝑥 times d𝑦 by d𝑥, that’s 𝑣 times
d𝑢 by d𝑥, minus 𝑢 times d𝑣 by d𝑥, which is 𝑦 times one. And, of course, that’s all over 𝑥
squared. We simplify this a little. And then we see that the derivative
of 𝑦 over 𝑥 is 𝑥 d𝑦 by d𝑥 minus 𝑦 over 𝑥 squared. So d two 𝑦 by d𝑥 squared is the
negative of this. We distribute our parentheses by
multiplying through by negative one. And we see that d two 𝑦 by d𝑥
squared is 𝑦 minus 𝑥 d𝑦 by d𝑥 over 𝑥 squared.
But, of course, we actually have an
expression for d𝑦 by d𝑥. It’s negative one minus 𝑦 over
𝑥. And so we replace d𝑦 by d𝑥 with
this. And we see that d two 𝑦 by d𝑥
squared is 𝑦 minus 𝑥 times negative one minus 𝑦 over 𝑥 all over 𝑥 squared. And then we distribute the
parentheses. Negative 𝑥 times negative one is
𝑥. And negative 𝑥 times negative 𝑦
over 𝑥, well the 𝑥’s cancel, and we get positive 𝑦. And so d two 𝑦 by d𝑥 squared is
two 𝑦 plus 𝑥 all over 𝑥 squared. We now have expressions for the
first and second derivative of 𝑦 with respect to 𝑥. So let’s substitute them into this
expression.
We get 𝑥 times two 𝑦 plus 𝑥 over
𝑥 squared plus two times negative one minus 𝑦 over 𝑥. We can see that we can cancel a
factor of 𝑥. And the first part of this becomes
two 𝑦 plus 𝑥 over 𝑥. It also then makes sense to
distribute our parentheses by multiplying each term by two. And so we get negative two minus
two 𝑦 over 𝑥. We’re actually gonna split our
first fraction up by dividing each bit by 𝑥. We get two 𝑦 over 𝑥 plus 𝑥 over
𝑥, which is just one. And now we see we have two 𝑦 over
𝑥 minus two 𝑦 over 𝑥, which is zero. And so this actually simplifies to
one minus two, which is negative one. And so 𝑥 times d two 𝑦 by d𝑥
squared plus two times d𝑦 by d𝑥 is negative one.
Now, it’s worth noting that we
could have originally divided both sides of our equation by negative two 𝑥. We then could’ve used the quotient
rule twice to find d𝑦 by d𝑥 and d two 𝑦 by d𝑥 squared. By dividing by negative two 𝑥, we
end up with an equation that’s explicitly in terms of 𝑥.
