### Video Transcript

Given that π₯ equals π‘ cubed plus five and π¦ equals π‘ squared minus three π‘, find π squared π¦ over ππ₯ squared.

Our question is actually asking us to find the second derivative π squared π¦ over ππ₯ squared. And in order to do this, what weβre gonna have to do first is actually find our first derivative. And to enable us to find ππ¦ ππ₯, weβre actually gonna have a look at the two equations we have here because actually what weβve got is a pair of parametric equations.

So therefore, in order to actually work out ππ¦ ππ₯ for our parametric equations, what weβre gonna do is we actually gonna apply the chain rule to help us find the first derivative. So what weβre gonna get is actually that ππ¦ ππ₯ is equal to ππ¦ ππ‘. And it is ππ‘ because thatβs the other variable in this particular question multiplied by ππ‘ ππ₯. And we do that because as you can see, weβre actually going to cancel out our ππ‘s because weβve got multiply by ππ‘ and then divide by ππ‘. So they cancel each other out to leave us with ππ¦ ππ₯. And weβre gonna get to that by actually differentiating each of our parametric equations one at a time.

So weβre gonna start first of all with π¦ equals π‘ squared minus three π‘. So we get that ππ¦ ππ‘ is equal to two π‘ minus three. And we go through the terms because if we got π‘ squared, differentiate that, we get two π‘ because you multiply the exponent by coefficient, so two by one, which gives us two. And then, we reduce the exponent by one. So we get two π‘ and then negative three π‘ just differentiates to negative three.

Okay, great, so weβve dealt with ππ¦ ππ‘. So now, letβs move on to our first parametric equation π₯ is equal to π‘ cubed plus five. So now, weβre going to differentiate this. And if we differentiate this, weβre actually gonna get ππ₯ ππ‘ is equal to three π‘ squared. Okay, great, however, if we look at our chain rule, we can actually see that oh! we donβt want the ππ₯ ππ‘, we actually want ππ‘ ππ₯.

So what we need to do now is find the reciprocal of ππ₯ ππ‘ so that we actually have ππ‘ ππ₯. And when we do that, we get that ππ‘ ππ₯ is equal to one over three π‘ squared. Okay, great, so now we found ππ¦ ππ‘. We found ππ‘ ππ₯. So now, we can actually apply the chain rule to find ππ¦ ππ₯. So we can say that therefore ππ¦ ππ₯ is equal to two π‘ minus three multiplied by one over three π‘ squared. So that is our ππ¦ ππ‘ multiplied by ππ‘ ππ₯, which is gonna give us two π‘ minus three over three π‘ squared.

So fantastic, weβve actually found our first derivative. We found ππ¦ ππ₯. Now, what we need to do is actually find our second derivative. And in order to find our second derivative, what weβre actually gonna do is weβre gonna find π ππ‘ of ππ¦ ππ₯. And then, weβre gonna multiply this by ππ‘ ππ₯.

And we do that because if we have a look, what we want is we want π squared π¦ over ππ₯ squared. And if we do it in this fashion, again, weβre actually gonna cancel out our π‘ terms because actually we only want it in terms of π₯ and π¦. So weβre gonna carry this out now to find our second derivative.

So what Iβm actually gonna do is Iβm actually gonna start by differentiating our ππ¦ ππ₯ β so two π‘ minus three over three π‘ squared with respect to π‘. So the quotient rule that weβre going to use says that if π¦ is equal to π’ over π£, ππ¦ ππ₯ is equal to π£ ππ’ ππ₯ minus π’ ππ£ ππ₯ over π£ squared. So now, weβre gonna apply this to actually find the derivative of our term. So we have our π’ is two π‘ minus three and our π£ is three π‘ squared.

So first of all, we want to actually find ππ’ ππ‘ and ππ£ ππ‘. So ππ’ ππ‘ is gonna be equal to just two because weβve actually differentiated two π‘ minus three, which just leaves us with two. And ππ£ ππ‘ is actually gonna give us six π‘. Okay, great, so now weβve got ππ’ ππ‘ and ππ£ ππ‘, letβs use the quotient rule and find out the differential with respect to π‘ of two π‘ minus three over three π‘ squared.

So therefore, what weβre gonna get is three π‘ squared multiplied by two because thatβs our π£ multiplied by our ππ’ ππ₯ minus two π‘ minus three multiplied by six π‘ because thatβs our π’ ππ£ ππ₯. And then, this is all divided by π£ squared. So itβs three π‘ squared all squared. So on the numerator, we have six π‘ squared minus 12π‘ squared plus 18π‘. And on the denominator, we have nine π‘ to the power of four. And we get that because three π‘ squared, well, we square three, we get nine. And then, π‘ to the power of two or π‘ squared squared is π‘ to the power of four. So great!

If we now simplify, weβre gonna get- well, six π‘ squared minus 12π‘ squared gives us negative six π‘ squared plus 18π‘ all over nine π‘ to the power of four. And then if we divide through by three π‘ β so weβre dividing the numerator by three π‘ and the denominator by three π‘ β weβre gonna get negative two π‘ plus six over three π‘ cubed which is gonna leave us with two then in parenthesis three minus π‘ because what Iβve done there is actually taking two as a factor of each of the terms in the numerator. And thatβs all over three π‘ cubed.

Okay, great, so weβve done that now and weβve actually found our π ππ‘ of two π‘ minus three over three π‘ squared using the quotient rule. So now, the final stage is that we need to multiply by ππ‘ ππ₯. And just as a quick reminder, ππ‘ ππ₯ was equal to one over three π‘ squared. Okay, so weβre gonna multiply that by our expression that we actually got from the previous step.

And if we do that, we multiply the numerators. So itβs two then in parenthesis three minus π‘ multiplied by one, which is gonna leave us with the same numerator. And then, our denominator is three π‘ cubed multiplied by three π‘ squared, which gives us nine π‘ to the power of five.

So therefore, we can say that given that π₯ equals π‘ cubed plus five and π¦ equals π‘ squared minus three π‘, the second differential is going to be equal to two and then open parenthesis three minus π‘ close parenthesis over nine π‘ to the power of five.