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

Given that 𝑥 = 3𝑡³ + 1 and 𝑦 = 5𝑡² − 𝑡, find 𝑑𝑦/𝑑𝑥.

03:19

### Video Transcript

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

Well, here we actually have a pair of parametric equations. And actually to find 𝑑𝑦 𝑑𝑥, we don’t have to write them in a Cartesian format. We can actually solve it using the chain rule. And the chain rule actually states that 𝑑𝑦 𝑑𝑥 is equal to 𝑑𝑦 and in this case 𝑑𝑡, because 𝑡 is our other variable multiplied by 𝑑𝑡 𝑑𝑥. And this is great because it enables us to actually differentiate each of our parametric equations separately and then combine them to find 𝑑𝑦 𝑑𝑥.

So let’s get on and do that. So first of all, I’m gonna start with 𝑦 equals five 𝑡 squared minus 𝑡. So to find 𝑑𝑦 𝑑𝑡, I’m gonna differentiate that. And my first term is just 10𝑡. And we get that because it’s two multiplied by five because that’s our exponent multiplied by a coefficient, which gives us 10. And that’s 𝑡 to the power of two minus one. And that’s because actually we subtract one from the exponent when we actually differentiate.

And then the second term differentiates two minus one. So great! We’ve got 10𝑡 minus one. So now we can actually move on and find the second part. But the second part as you can see, as we’re trying to find 𝑑𝑡 𝑑𝑥, we’re gonna have to do that in a slightly different way. And the way we do that is we actually find 𝑑𝑥 𝑑𝑡 and then find the reciprocal of that because that will give us 𝑑𝑡 𝑑𝑥.

Okay, so let’s get on and find that now. So we have 𝑥 is equal to three 𝑡 cubed plus one. So therefore we’re gonna get 𝑑𝑥 𝑑𝑡 is equal to 9𝑡 squared. Again and we got that because we have three multiplied by three cause it’s the exponent multiplied by the coefficient. And that’s 𝑡 to the power of three subtract one. And that’s that because we actually subtract one when we’re differentiating.

So we’ve got 9𝑡 squared. And positive one just differentiates to zero. So great! We’ve got 9𝑡 squared for 𝑑𝑥 𝑑𝑡. So now we can say that, therefore, 𝑑𝑡 𝑑𝑥 is equal to one over 9𝑡 squared. That’s because as we said before, we find a reciprocal of 9𝑡 squared.

So great! We’ve got 𝑑𝑦 𝑑𝑡 and 𝑑𝑡 𝑑𝑥. Let’s use the chain rule to find 𝑑𝑦 𝑑𝑥. So therefore using the chain rule, we can say that 𝑑𝑦 𝑑𝑥 is equal to 10𝑡 minus one multiplied by one over 9𝑡 squared. So therefore we can say that 𝑑𝑦 𝑑𝑥 is equal to 10𝑡 minus one over 9𝑡 squared, given that 𝑥 is equal to three 𝑡 cubed plus one and 𝑦 is equal to five 𝑡 squared minus 𝑡.

And now we’ve got the final answer. I just wanna do a quick recap to remind us of what we’ve done. So first of all, we can see that actually we have parametric equations. So we’re gonna use the chain rule to find 𝑑𝑦 𝑑𝑥. And then we actually differentiate both of our parametric equations. And that gives us in this case 𝑑𝑦 𝑑𝑡 and then 𝑑𝑥 𝑑𝑡. But remembering that we actually had to find the reciprocal of 𝑑𝑥 𝑑𝑡 to find 𝑑𝑡 𝑑𝑥, then finally we apply the chain rule to get our final answer.