### Video Transcript

Given that π¦ equals ππ₯ cubed plus ππ₯ squared, the third-order derivative of π¦ is equal to negative 18, and π two π¦ over ππ₯ squared when π₯ is equal to two is equal to negative 14, find π and π.

So the first thing we need to do in this question is actually find our first-, second-, and third-order derivatives. So weβre gonna start with our first-order derivative. And just to remind us how we actually differentiate if we have a function in π₯, then if itβs in the form ππ₯ to the power of π, then our first-order derivative is gonna be equal to πππ₯ to the power of π minus one. So you multiply the coefficient by the exponent and reduce the exponent by one.

Okay, so letβs use that and letβs find the first-order derivative of our function. So the first term is gonna be three ππ₯ squared. And thatβs because we multiplied our exponent three by a coefficient π. So we get three π and then weβve reduced the exponent by one. So we get π₯ squared. So then, the second term is just plus two ππ₯. Again, weβve differentiated that term. Okay, so thatβs our first-order derivative.

What weβre gonna move onto now is going to differentiate again to find our second-order derivative. So our second-order derivative which Iβve shown here with π¦ prime prime. And all as weβve got it in the question π two π¦ over ππ₯ squared is gonna be six ππ₯ plus two π. So again, we just differentiate that in the normal way. So we have three π multiplied by two β our coefficient multiplied by our exponent β and then π₯ to the power of two minus one, which is just π₯ β so π₯ to the power of one. Okay, great, so now weβve found our second-order derivative.

So now, great, we can actually move on and find our third-order derivative. And all we do to find this is actually differentiate again, which is just gonna leave us with six π because if we differentiate six ππ₯, weβre just left with six π. And if we differentiate plus two π, then itβs just going to be zero because if we differentiate any number that hasnβt got an π₯ term in, it just goes to zero. Okay, so now weβve done that β weβve got our first-, second-, and third-order derivatives β letβs see what we do next.

Well, letβs start with this bit of information. We know that our third-order derivative is gonna be equal to negative 18. So we can now set up an equation. And we can actually set up that new equation because we can say that our third-order derivative is equal to negative 18. So can sub that in. So we get negative 18 is equal to six π. Okay, so therefore, we can say that π is going to be equal to negative three. And weβve got that because weβve actually divided through by six. And also, we just flipped it just so we got the π on the left-hand side.

So fantastic, weβve now found π. So now, what we need to do is move on and have a go at finding π. And in order to do that, what weβre gonna use is this bit of information here. Because if we take a look at this bit of information, weβve got that π two π¦ over ππ₯ squared β so our second-order derivative β is equal to negative 14 when π₯ is equal to two. So therefore, in order to actually find π, what we can do is substitute in π₯ is equal to two and π¦ prime prime is equal to negative 14. And this is gonna give us that negative 14 is equal to six π multiplied by two plus two π.

But weβve now got something else. Weβre gonna substitute in because we know π. So we can substitute in π is equal to negative three. So therefore, we get that negative 14 equals six multiplied by negative three multiplied by two plus two π. So we have negative 14 equals negative 36 plus two π. So then, what we do is we actually add 36 to each side. So we have 22 equals two π. So therefore, if we divide by two, we get π is equal to 11.

So that means that we can say that given that π¦ equals ππ₯ cubed plus ππ₯ squared which third-order derivative of π¦ is equal to negative 18 and π two π¦ over ππ₯ squared when π₯ is equal to two is equal to negative 14, π is equal to negative three and π is equal to 11.