 Let us find y double prime if the relationship between x and y is given by the equation x to the fourth plus y to the fourth equals 16. So let's unwrap this a little bit here. What does y double prime even mean? This means the second derivative. So y double prime is short for d squared y over dx squared. What's with the squares going on here? This second derivative means that we take the first derivative dy over dx and we take its derivative with respect to x right here. This right here is why the d gets squared, right? So you have the d showing up twice and the dx gets squared because it shows up twice as well. The second derivative is then the derivative of the derivative. So we're taking dy prime of dx, right? So if you want to compute the second derivative of a function, it just means take the derivative with respect to x of the original variable with respect to x. We take the derivative twice. So if you want to find the second derivative, it means we're going to take the derivative twice. But when you look at this relationship x to the fourth plus y to the fourth equals 16, this is not an explicit relationship. We don't have y equals some function f of x. We have that x and y are sort of commingling on their same side of the equation. That's okay. We can compute the derivative implicitly as well. We also take the second derivative implicitly. So to find the first derivative y prime, we're just going to take the derivative of both sides, d dx of x to the fourth plus y to the fourth. This is equal to d dx of 16. Now on the right hand side, we're going to tackle that one first. The derivative of constant is going to equal zero. So that's going to disappear. On the left hand side, we're just going to take the derivative in the usual fashion, right? x to the fourth was one of the fourth prime is equal to zero here. Because we're taking the derivative of the sum, we can take the derivative of the sum and separately. So we take x to the fourth prime plus y to the fourth prime. This is equal to zero. Now the thing to remember is we're taking the derivative with respect to x. So with respect to x, the derivative of x to the fourth by the power rule will be 4x cubed. Now we take the derivative of y to the fourth. Since we're taking the derivative with respect to x, the power rule does apply. We're going to get 4y cubed. But the chain rule also applies. We need to take the derivative of y. y prime here is this dy over dx. This is all equal to zero. Now in order to find the second derivative, we could take the derivative at this moment, but it will be profitable to us to solve for y prime first and then take the derivative. So let's see how that would work. If we're going to solve for y prime, we would subtract 4x cubed from both sides of the equation. This gives us a 4y cubed y prime is equal to negative 4x cubed. Then we divide both sides by the coefficient 4y cubed. Do that to both sides. You'll notice that the 4 can cancel out right here. And so we see that y prime is going to equal a negative x cubed over y cubed. That's a cute little derivative right there. It's not so bad. That's not the final answer though. We're looking for y double prime. So how do we find the second derivative? Well, to find the second derivative, we need to take the derivative of the derivative. So if we have y prime on the left hand side, we're going to take the derivative with respect to x. On the right hand side, we have to take the derivative with respect to x of this function right here, negative x cubed over y cubed. For which on the left hand side, if you take the derivative of the derivative, use the second derivative so we could just call that y double prime. On the right hand side, let's calculate the derivative again implicitly. We can factor out the negative sign. So we have x cubed over y cubed. So we're going to take the derivative. If you wanted to, you could also treat this as in a slightly different manner. You could take x over y cubed prime. So if you wanted to, you could do those various ways. Neither of these avoid the quotient rule really. So I think I'm going to stick with my first approach right here. Using the quotient rule, we're going to get, by the usual quotient rule, low d high, the derivative of x cubed will be 3x squared, minus high d low 3y squared y prime. We need to remember the y prime there because we're taking the derivative with respect to x. And the denominator, we're going to get y cubed and we're going to square it. So let's try to simplify this expression on the right hand side for the second derivative. We're going to get, what do we have going on right here? We see that in the numerator, we have, let's actually try to factor out a common divisor, a common denominator, excuse me, a common factor, a common divisor here. So you can see that both can offer an x squared. This one, we can get an x squared times an x. We can also grab a 3. That's common to both. So in terms of the y squared, we can get a y squared, and this one can be a y squared times a y. So it turns out there is some stuff we can factor out. We're going to then factor these expressions out in the numerator. We took out a 3x squared y squared that leaves behind a y for the first term, right? So we get the y right there. We got rid of the 3x squared. Then the next part, we're going to take away the x squared, so we're left with an x. We got rid of the 3, the y squared, and we're left with y prime. And this all sits above y to the sixth. The y cubed squared would be the sixth power. This leads to a simplification. The y squared we took out of the top can simplify to be a y to the fourth right here. And this looks pretty good, right? 3x squared times y minus x, y prime over x to the fourth. You'll notice, wait, when I had the derivative of y, it was a formula of x and y. The second derivative is a formula of x, y, and y prime. Turns out we don't need y prime because like as we observed earlier, y prime is just negative x cubed over y cubed. We could substitute that in for y prime in order to help us simplify this thing. Let's do that very thing. We get negative 3x squared times y minus x times. Remember the derivative, we got negative x cubed over y cubed, and this sits above the y to the fourth. Like so. Alright, so what's going on here? We see that we're going to have a double negative, so this becomes a positive. We're going to get an x to the fourth. So we have negative 3x squared times y plus x to the fourth over y cubed all over y to the fourth. Now, I'm not a big fan of fractions inside of fractions. I want to get rid of that y cube that's in the numerator. So I'm going to multiply the big fraction by y cube, but I have to do it to the denominator as well so that things are proportional. So we're going to distribute the y cubed across that sum right there. That then produces for us 3x squared times. When we times the y by y cube, we're going to get y to the fourth, and then we're going to get x to the fourth. And then the denominator, we get a y to the seventh, y to the fourth times y cubed. And so there we go. That gives us then the derivative that we were seeking, right? Maybe what's left to say here is it turns out there's one more simplification we can do. We got rid of all the y primes, but one cool simplification I want to mention here is let's zoom out a little bit. Let's go back to the original problem. Remember, y double prime equals x squared, or x to the fourth plus y to the fourth is equal to 16, right? You'll notice that here, x to the fourth plus y to the fourth is equal to 16. Let's go back to our second derivative. You'll notice in this formula it's like, wait a second, y to the fourth plus x to the fourth. I've seen that before. The original equation, that's actually equal to 16. So you'll notice that at this moment we plugged in the first derivative to help simplify it. We can also plug in the so-called zeroth derivative, that is the original equation, can be plugged in here as well. This is equal to 16. So we end up with y double prime is equal to, there should be a negative sign here that didn't carry forward, negative 3x squared times 16 all over y to the seventh. 16 times 3 is a 48. And so we see the second derivative is negative 48x squared over y to the seventh. And so this is the key that you need to remember when it comes to calculating the second derivative implicitly, is that to calculate the second derivative, you just take the derivative and then calculate the derivative again. Take the derivative of the derivative. You will need to substitute in the formula you have for the first derivative to help simplify the second derivative. And in some situations like this example, you actually use the original equation to help simplify it as well. Don't expect that to happen so often, but I actually like this example because it's kind of this fun little thing to happen. We can simplify the x to the fourth plus y to the fourth with the 16.