# Video: Finding the Second Derivative of a Function Defined by Parametric Equations

Given that 𝑥 = 𝑡³ + 5 and 𝑦 = 𝑡² − 3𝑡, find 𝑑²𝑦/𝑑𝑥².

07:25

### Video Transcript

Given that 𝑥 equals 𝑡 cubed plus five and 𝑦 equals 𝑡 squared minus three 𝑡, find 𝑑 squared 𝑦 over 𝑑𝑥 squared.

Our question is actually asking us to find the second derivative 𝑑 squared 𝑦 over 𝑑𝑥 squared. And in order to do this, what we’re gonna have to do first is actually find our first derivative. And to enable us to find 𝑑𝑦 𝑑𝑥, we’re actually gonna have a look at the two equations we have here because actually what we’ve got is a pair of parametric equations.

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

So we’re gonna start first of all with 𝑦 equals 𝑡 squared minus three 𝑡. So we get that 𝑑𝑦 𝑑𝑡 is equal to two 𝑡 minus three. And we go through the terms because if we got 𝑡 squared, differentiate that, we get two 𝑡 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 𝑡 and then negative three 𝑡 just differentiates to negative three.

Okay, great, so we’ve dealt with 𝑑𝑦 𝑑𝑡. So now, let’s move on to our first parametric equation 𝑥 is equal to 𝑡 cubed plus five. So now, we’re going to differentiate this. And if we differentiate this, we’re actually gonna get 𝑑𝑥 𝑑𝑡 is equal to three 𝑡 squared. Okay, great, however, if we look at our chain rule, we can actually see that oh! we don’t want the 𝑑𝑥 𝑑𝑡, we actually want 𝑑𝑡 𝑑𝑥.

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

So fantastic, we’ve actually found our first derivative. We found 𝑑𝑦 𝑑𝑥. Now, what we need to do is actually find our second derivative. And in order to find our second derivative, what we’re actually gonna do is we’re gonna find 𝑑 𝑑𝑡 of 𝑑𝑦 𝑑𝑥. And then, we’re gonna multiply this by 𝑑𝑡 𝑑𝑥.

And we do that because if we have a look, what we want is we want 𝑑 squared 𝑦 over 𝑑𝑥 squared. And if we do it in this fashion, again, we’re actually gonna cancel out our 𝑡 terms because actually we only want it in terms of 𝑥 and 𝑦. So we’re gonna carry this out now to find our second derivative.

So what I’m actually gonna do is I’m actually gonna start by differentiating our 𝑑𝑦 𝑑𝑥 — so two 𝑡 minus three over three 𝑡 squared with respect to 𝑡. So the quotient rule that we’re going to use says that if 𝑦 is equal to 𝑢 over 𝑣, 𝑑𝑦 𝑑𝑥 is equal to 𝑣 𝑑𝑢 𝑑𝑥 minus 𝑢 𝑑𝑣 𝑑𝑥 over 𝑣 squared. So now, we’re gonna apply this to actually find the derivative of our term. So we have our 𝑢 is two 𝑡 minus three and our 𝑣 is three 𝑡 squared.

So first of all, we want to actually find 𝑑𝑢 𝑑𝑡 and 𝑑𝑣 𝑑𝑡. So 𝑑𝑢 𝑑𝑡 is gonna be equal to just two because we’ve actually differentiated two 𝑡 minus three, which just leaves us with two. And 𝑑𝑣 𝑑𝑡 is actually gonna give us six 𝑡. Okay, great, so now we’ve got 𝑑𝑢 𝑑𝑡 and 𝑑𝑣 𝑑𝑡, let’s use the quotient rule and find out the differential with respect to 𝑡 of two 𝑡 minus three over three 𝑡 squared.

So therefore, what we’re gonna get is three 𝑡 squared multiplied by two because that’s our 𝑣 multiplied by our 𝑑𝑢 𝑑𝑥 minus two 𝑡 minus three multiplied by six 𝑡 because that’s our 𝑢 𝑑𝑣 𝑑𝑥. And then, this is all divided by 𝑣 squared. So it’s three 𝑡 squared all squared. So on the numerator, we have six 𝑡 squared minus 12𝑡 squared plus 18𝑡. And on the denominator, we have nine 𝑡 to the power of four. And we get that because three 𝑡 squared, well, we square three, we get nine. And then, 𝑡 to the power of two or 𝑡 squared squared is 𝑡 to the power of four. So great!

If we now simplify, we’re gonna get- well, six 𝑡 squared minus 12𝑡 squared gives us negative six 𝑡 squared plus 18𝑡 all over nine 𝑡 to the power of four. And then if we divide through by three 𝑡 — so we’re dividing the numerator by three 𝑡 and the denominator by three 𝑡 — we’re gonna get negative two 𝑡 plus six over three 𝑡 cubed which is gonna leave us with two then in parenthesis three minus 𝑡 because what I’ve done there is actually taking two as a factor of each of the terms in the numerator. And that’s all over three 𝑡 cubed.

Okay, great, so we’ve done that now and we’ve actually found our 𝑑 𝑑𝑡 of two 𝑡 minus three over three 𝑡 squared using the quotient rule. So now, the final stage is that we need to multiply by 𝑑𝑡 𝑑𝑥. And just as a quick reminder, 𝑑𝑡 𝑑𝑥 was equal to one over three 𝑡 squared. Okay, so we’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 it’s two then in parenthesis three minus 𝑡 multiplied by one, which is gonna leave us with the same numerator. And then, our denominator is three 𝑡 cubed multiplied by three 𝑡 squared, which gives us nine 𝑡 to the power of five.

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