# Question Video: Finding the Second Derivative of a Function Defined by Parametric Equations Mathematics • Higher Education

Given that π₯ = π‘Β³ + 5 and π¦ = π‘Β² β 3π‘, find πΒ²π¦/ππ₯Β².

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### Video Transcript

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

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

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

So weβre gonna start first of all with π¦ equals π‘ squared minus three π‘. So we get that ππ¦ ππ‘ 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 weβve dealt with ππ¦ ππ‘. So now, letβs move on to our first parametric equation π₯ is equal to π‘ cubed plus five. So now, weβre going to differentiate this. And if we differentiate this, weβre actually gonna get ππ₯ ππ‘ is equal to three π‘ squared. Okay, great, however, if we look at our chain rule, we can actually see that oh! we donβt want the ππ₯ ππ‘, we actually want ππ‘ ππ₯.

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

So fantastic, weβve actually found our first derivative. We found ππ¦ ππ₯. Now, what we need to do is actually find our second derivative. And in order to find our second derivative, what weβre actually gonna do is weβre gonna find π ππ‘ of ππ¦ ππ₯. And then, weβre gonna multiply this by ππ‘ ππ₯.

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

So what Iβm actually gonna do is Iβm actually gonna start by differentiating our ππ¦ ππ₯ β so two π‘ minus three over three π‘ squared with respect to π‘. So the quotient rule that weβre going to use says that if π¦ is equal to π’ over π£, ππ¦ ππ₯ is equal to π£ ππ’ ππ₯ minus π’ ππ£ ππ₯ over π£ squared. So now, weβre 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 ππ’ ππ‘ and ππ£ ππ‘. So ππ’ ππ‘ is gonna be equal to just two because weβve actually differentiated two π‘ minus three, which just leaves us with two. And ππ£ ππ‘ is actually gonna give us six π‘. Okay, great, so now weβve got ππ’ ππ‘ and ππ£ ππ‘, letβs 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 weβre gonna get is three π‘ squared multiplied by two because thatβs our π£ multiplied by our ππ’ ππ₯ minus two π‘ minus three multiplied by six π‘ because thatβs our π’ ππ£ ππ₯. And then, this is all divided by π£ squared. So itβs 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, weβre 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 weβre dividing the numerator by three π‘ and the denominator by three π‘ β weβre 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 Iβve done there is actually taking two as a factor of each of the terms in the numerator. And thatβs all over three π‘ cubed.

Okay, great, so weβve done that now and weβve actually found our π ππ‘ 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 ππ‘ ππ₯. And just as a quick reminder, ππ‘ ππ₯ was equal to one over three π‘ squared. Okay, so weβre gonna multiply that by our expression that we actually got from the previous step.

And if we do that, we multiply the numerators. So itβs 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.