### Video Transcript

Given that π₯ squared plus three π¦ squared equals three, determine π¦ double prime by implicit differentiation.

This π¦ double prime is the second derivative of π¦ with respect to π₯. And weβre told to find it by implicit differentiation β that is by differentiating both sides of our equation with respect to π₯. So letβs do this. π by ππ₯ all of π₯ squared plus three π¦ squared is then equal to π by ππ₯ of three. And we can differentiate the two terms on the left-hand side separately.

π by ππ₯ of π₯ squared is straightforward. Itβs just two π₯. π by ππ₯ of three π¦ squared though is more difficult. We have to use the chain rule. ππ by ππ₯ is ππ by ππ¦ times ππ¦ by ππ₯. So with π equal to three π¦ squared, this is π by ππ¦ of three π¦ squared times ππ¦ by ππ₯. And π by ππ¦ of three π¦ squared is six π¦. So π by ππ₯ of three π¦ squared is six π¦ times ππ¦ by ππ₯ on the right-hand side. π by ππ₯ of three is zero. And so putting it altogether, we have that two π₯ plus six π¦ times ππ¦ by ππ₯ is zero. And we can rearrange this to find ππ¦ by ππ₯.

First, subtracting two π₯ from both sides and then dividing by six π¦ noticing that we can cancel a factor of two to get minus π₯ over three π¦. And perhaps, we should follow the notation in our problem and use primes to represent derivatives. Then, ππ¦ by ππ₯ is π¦ prime. So what π¦ prime is minus π₯ over three π¦. But what is π¦ double prime? Well, we need to differentiate again with respect to π₯. Differentiating both sides with respect to π₯, we get π by ππ₯ of π¦ prime equals π by ππ₯ of minus π₯ over three π¦. And π by ππ₯ of π¦ prime is π by ππ₯ of the first derivative of π¦ with respect to π₯, which is therefore the second derivative of π¦ with respect to π₯. And thatβs the π¦ double prime that weβre looking for.

On the right-hand side, weβve got the derivative of a quotient. And so weβre going to need the quotient rule and here it is in all its glory. Now applying it with π of π₯ equal to π₯ and π of π₯ equal to three π¦, not forgetting the minus sign, we get minus three π¦ times the derivative of π₯ with respect to π₯ minus π₯ times the derivative of three π¦ with respect to π₯ all over three π¦ squared.

Now we can simplify. The derivative of π₯ with respect to π₯ is just one. And so the first term in the numerator becomes three π¦. The derivative over three π¦ with respect to π₯ is three times the derivative of π¦ with respect to π₯ β three π¦ prime. And so the second term becomes three π₯ π¦ prime. And finally, the denominator is nine π¦ squared.

We see that we can simplify further by cancelling a factor of three in the numerator and denominator. And so we get minus π¦ minus π₯ π¦ prime over three π¦ squared. We can also absorb the minus sign into the fraction. And so we see that π¦ double prime is π₯ times π¦ prime minus π¦ all over three π¦ squared. Now, weβve written π¦ double prime in terms of π₯, π¦, and π¦ prime the first derivative of π¦.

But really, weβd prefer to have π¦ double prime written in terms of π₯ and π¦ alone. That way if weβre given a point with coordinates π₯ and π¦ which lies on our curve, we can just substitute those coordinates for π₯ and π¦ in our expression to find the value of π¦ double prime at that point. To get π¦ double prime written in terms of π₯ and π¦ alone, we need to get rid of this π¦ prime somehow. How do we do that? Well, we can substitute the expression we have for π¦ prime in terms of π₯ and π¦.

Making this substitution, we get π¦ double prime in terms of π₯ and π¦. And we now just have to simplify. We multiply both numerator and denominator by three π¦. So the first term in the numerator becomes minus π₯ squared and the second term becomes minus three π¦ squared. The denominator becomes nine π¦ cubed. We can pull out a minus sign here. But looking at this expression alone, it looks like weβre done.

But have a look at that numerator π₯ squared plus three π¦ squared. Weβre told in our problem that π₯ squared plus three π¦ squared is three. And so we get an unexpected simplification: π¦ double prime equals negative three over nine π¦ cubed. And we can cancel out another common factor of three. And we get our final answer that π¦ double prime the second derivative of π¦ with respect to π₯ is negative one over three π¦ cubed.

Here we used implicit differentiation twice first to find π¦ prime in terms of π₯ and π¦ and then again to find π¦ double prime in terms of π₯, π¦, and π¦ prime. Substituting the expression for π¦ prime in terms of π₯ and π¦, we got π¦ double prime in terms of π₯ and π¦ alone. And then, we noticed that thereβs really unexpected and almost magical substitution that we could do from the question, which allowed us to write π¦ double prime in terms of π¦ alone.