Video: Finding the Second Derivative of a Function Defined Implicitly Using Implicit Differentiation

Given that π‘₯Β² + 3𝑦² = 3, determine 𝑦″ by implicit differentiation.

05:57

Video Transcript

Given that π‘₯ 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 π‘₯. And we’re told to find it by implicit differentiation β€” that is by differentiating both sides of our equation with respect to π‘₯. So let’s do this. 𝑑 by 𝑑π‘₯ all of π‘₯ squared plus three 𝑦 squared is then equal to 𝑑 by 𝑑π‘₯ of three. And we can differentiate the two terms on the left-hand side separately.

𝑑 by 𝑑π‘₯ of π‘₯ squared is straightforward. It’s just two π‘₯. 𝑑 by 𝑑π‘₯ of three 𝑦 squared though is more difficult. We have to use the chain rule. 𝑑𝑓 by 𝑑π‘₯ is 𝑑𝑓 by 𝑑𝑦 times 𝑑𝑦 by 𝑑π‘₯. So with 𝑓 equal to three 𝑦 squared, this is 𝑑 by 𝑑𝑦 of three 𝑦 squared times 𝑑𝑦 by 𝑑π‘₯. And 𝑑 by 𝑑𝑦 of three 𝑦 squared is six 𝑦. So 𝑑 by 𝑑π‘₯ of three 𝑦 squared is six 𝑦 times 𝑑𝑦 by 𝑑π‘₯ on the right-hand side. 𝑑 by 𝑑π‘₯ of three is zero. And so putting it altogether, we have that two π‘₯ plus six 𝑦 times 𝑑𝑦 by 𝑑π‘₯ is zero. And we can rearrange this to find 𝑑𝑦 by 𝑑π‘₯.

First, subtracting two π‘₯ from both sides and then dividing by six 𝑦 noticing that we can cancel a factor of two to get minus π‘₯ over three 𝑦. And perhaps, we should follow the notation in our problem and use primes to represent derivatives. Then, 𝑑𝑦 by 𝑑π‘₯ is 𝑦 prime. So what 𝑦 prime is minus π‘₯ over three 𝑦. But what is 𝑦 double prime? Well, we need to differentiate again with respect to π‘₯. Differentiating both sides with respect to π‘₯, we get 𝑑 by 𝑑π‘₯ of 𝑦 prime equals 𝑑 by 𝑑π‘₯ of minus π‘₯ over three 𝑦. And 𝑑 by 𝑑π‘₯ of 𝑦 prime is 𝑑 by 𝑑π‘₯ of the first derivative of 𝑦 with respect to π‘₯, which is therefore the second derivative of 𝑦 with respect to π‘₯. 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 π‘₯ equal to π‘₯ and 𝑑 of π‘₯ equal to three 𝑦, not forgetting the minus sign, we get minus three 𝑦 times the derivative of π‘₯ with respect to π‘₯ minus π‘₯ times the derivative of three 𝑦 with respect to π‘₯ all over three 𝑦 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 𝑦. The derivative over three 𝑦 with respect to π‘₯ is three times the derivative of 𝑦 with respect to π‘₯ β€” three 𝑦 prime. And so the second term becomes three π‘₯ 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 π‘₯ 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 π‘₯ times 𝑦 prime minus 𝑦 all over three 𝑦 squared. Now, we’ve written 𝑦 double prime in terms of π‘₯, 𝑦, and 𝑦 prime the first derivative of 𝑦.

But really, we’d prefer to have 𝑦 double prime written in terms of π‘₯ and 𝑦 alone. That way if we’re given a point with coordinates π‘₯ and 𝑦 which lies on our curve, we can just substitute those coordinates for π‘₯ 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 π‘₯ 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 π‘₯ and 𝑦.

Making this substitution, we get 𝑦 double prime in terms of π‘₯ 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 π‘₯ 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 π‘₯ squared plus three 𝑦 squared. We’re told in our problem that π‘₯ 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 π‘₯ is negative one over three 𝑦 cubed.

Here we used implicit differentiation twice first to find 𝑦 prime in terms of π‘₯ and 𝑦 and then again to find 𝑦 double prime in terms of π‘₯, 𝑦, and 𝑦 prime. Substituting the expression for 𝑦 prime in terms of π‘₯ and 𝑦, we got 𝑦 double prime in terms of π‘₯ 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.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.