Given that 𝑥 squared plus three 𝑦 squared equals three, determine 𝑦 double prime by implicit differentiation.
This 𝑦 double prime is the second derivative of 𝑦 with respect to 𝑥. And we’re told to find it by implicit differentiation — that is by differentiating both sides of our equation with respect to 𝑥. So let’s do this. 𝑑 by 𝑑𝑥 all of 𝑥 squared plus three 𝑦 squared is then equal to 𝑑 by 𝑑𝑥 of three. And we can differentiate the two terms on the left-hand side separately.
𝑑 by 𝑑𝑥 of 𝑥 squared is straightforward. It’s just two 𝑥. 𝑑 by 𝑑𝑥 of three 𝑦 squared though is more difficult. We have to use the chain rule. 𝑑𝑓 by 𝑑𝑥 is 𝑑𝑓 by 𝑑𝑦 times 𝑑𝑦 by 𝑑𝑥. So with 𝑓 equal to three 𝑦 squared, this is 𝑑 by 𝑑𝑦 of three 𝑦 squared times 𝑑𝑦 by 𝑑𝑥. And 𝑑 by 𝑑𝑦 of three 𝑦 squared is six 𝑦. So 𝑑 by 𝑑𝑥 of three 𝑦 squared is six 𝑦 times 𝑑𝑦 by 𝑑𝑥 on the right-hand side. 𝑑 by 𝑑𝑥 of three is zero. And so putting it altogether, we have that two 𝑥 plus six 𝑦 times 𝑑𝑦 by 𝑑𝑥 is zero. And we can rearrange this to find 𝑑𝑦 by 𝑑𝑥.
First, subtracting two 𝑥 from both sides and then dividing by six 𝑦 noticing that we can cancel a factor of two to get minus 𝑥 over three 𝑦. And perhaps, we should follow the notation in our problem and use primes to represent derivatives. Then, 𝑑𝑦 by 𝑑𝑥 is 𝑦 prime. So what 𝑦 prime is minus 𝑥 over three 𝑦. But what is 𝑦 double prime? Well, we need to differentiate again with respect to 𝑥. Differentiating both sides with respect to 𝑥, we get 𝑑 by 𝑑𝑥 of 𝑦 prime equals 𝑑 by 𝑑𝑥 of minus 𝑥 over three 𝑦. And 𝑑 by 𝑑𝑥 of 𝑦 prime is 𝑑 by 𝑑𝑥 of the first derivative of 𝑦 with respect to 𝑥, which is therefore the second derivative of 𝑦 with respect to 𝑥. And that’s the 𝑦 double prime that we’re looking for.
On the right-hand side, we’ve got the derivative of a quotient. And so we’re going to need the quotient rule and here it is in all its glory. Now applying it with 𝑓 of 𝑥 equal to 𝑥 and 𝑑 of 𝑥 equal to three 𝑦, not forgetting the minus sign, we get minus three 𝑦 times the derivative of 𝑥 with respect to 𝑥 minus 𝑥 times the derivative of three 𝑦 with respect to 𝑥 all over three 𝑦 squared.
Now we can simplify. The derivative of 𝑥 with respect to 𝑥 is just one. And so the first term in the numerator becomes three 𝑦. The derivative over three 𝑦 with respect to 𝑥 is three times the derivative of 𝑦 with respect to 𝑥 — three 𝑦 prime. And so the second term becomes three 𝑥 𝑦 prime. And finally, the denominator is nine 𝑦 squared.
We see that we can simplify further by cancelling a factor of three in the numerator and denominator. And so we get minus 𝑦 minus 𝑥 𝑦 prime over three 𝑦 squared. We can also absorb the minus sign into the fraction. And so we see that 𝑦 double prime is 𝑥 times 𝑦 prime minus 𝑦 all over three 𝑦 squared. Now, we’ve written 𝑦 double prime in terms of 𝑥, 𝑦, and 𝑦 prime the first derivative of 𝑦.
But really, we’d prefer to have 𝑦 double prime written in terms of 𝑥 and 𝑦 alone. That way if we’re given a point with coordinates 𝑥 and 𝑦 which lies on our curve, we can just substitute those coordinates for 𝑥 and 𝑦 in our expression to find the value of 𝑦 double prime at that point. To get 𝑦 double prime written in terms of 𝑥 and 𝑦 alone, we need to get rid of this 𝑦 prime somehow. How do we do that? Well, we can substitute the expression we have for 𝑦 prime in terms of 𝑥 and 𝑦.
Making this substitution, we get 𝑦 double prime in terms of 𝑥 and 𝑦. And we now just have to simplify. We multiply both numerator and denominator by three 𝑦. So the first term in the numerator becomes minus 𝑥 squared and the second term becomes minus three 𝑦 squared. The denominator becomes nine 𝑦 cubed. We can pull out a minus sign here. But looking at this expression alone, it looks like we’re done.
But have a look at that numerator 𝑥 squared plus three 𝑦 squared. We’re told in our problem that 𝑥 squared plus three 𝑦 squared is three. And so we get an unexpected simplification: 𝑦 double prime equals negative three over nine 𝑦 cubed. And we can cancel out another common factor of three. And we get our final answer that 𝑦 double prime the second derivative of 𝑦 with respect to 𝑥 is negative one over three 𝑦 cubed.
Here we used implicit differentiation twice first to find 𝑦 prime in terms of 𝑥 and 𝑦 and then again to find 𝑦 double prime in terms of 𝑥, 𝑦, and 𝑦 prime. Substituting the expression for 𝑦 prime in terms of 𝑥 and 𝑦, we got 𝑦 double prime in terms of 𝑥 and 𝑦 alone. And then, we noticed that there’s really unexpected and almost magical substitution that we could do from the question, which allowed us to write 𝑦 double prime in terms of 𝑦 alone.