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

Given that π‘₯ = 3𝑑³ + 1 and 𝑦 = 5𝑑² βˆ’ 𝑑, find 𝑑𝑦/𝑑π‘₯.


Video Transcript

Given that π‘₯ equals three 𝑑 cubed plus one and 𝑦 equals five 𝑑 squared minus 𝑑, find 𝑑𝑦 𝑑π‘₯.

Well, here we actually have a pair of parametric equations. And actually to find 𝑑𝑦 𝑑π‘₯, 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 𝑑π‘₯ is equal to 𝑑𝑦 and in this case 𝑑𝑑, because 𝑑 is our other variable multiplied by 𝑑𝑑 𝑑π‘₯. And this is great because it enables us to actually differentiate each of our parametric equations separately and then combine them to find 𝑑𝑦 𝑑π‘₯.

So let’s get on and do that. So first of all, I’m gonna start with 𝑦 equals five 𝑑 squared minus 𝑑. So to find 𝑑𝑦 𝑑𝑑, I’m gonna differentiate that. And my first term is just 10𝑑. 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 𝑑 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 10𝑑 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 𝑑𝑑 𝑑π‘₯, we’re gonna have to do that in a slightly different way. And the way we do that is we actually find 𝑑π‘₯ 𝑑𝑑 and then find the reciprocal of that because that will give us 𝑑𝑑 𝑑π‘₯.

Okay, so let’s get on and find that now. So we have π‘₯ is equal to three 𝑑 cubed plus one. So therefore we’re gonna get 𝑑π‘₯ 𝑑𝑑 is equal to 9𝑑 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 𝑑 to the power of three subtract one. And that’s that because we actually subtract one when we’re differentiating.

So we’ve got 9𝑑 squared. And positive one just differentiates to zero. So great! We’ve got 9𝑑 squared for 𝑑π‘₯ 𝑑𝑑. So now we can say that, therefore, 𝑑𝑑 𝑑π‘₯ is equal to one over 9𝑑 squared. That’s because as we said before, we find a reciprocal of 9𝑑 squared.

So great! We’ve got 𝑑𝑦 𝑑𝑑 and 𝑑𝑑 𝑑π‘₯. Let’s use the chain rule to find 𝑑𝑦 𝑑π‘₯. So therefore using the chain rule, we can say that 𝑑𝑦 𝑑π‘₯ is equal to 10𝑑 minus one multiplied by one over 9𝑑 squared. So therefore we can say that 𝑑𝑦 𝑑π‘₯ is equal to 10𝑑 minus one over 9𝑑 squared, given that π‘₯ is equal to three 𝑑 cubed plus one and 𝑦 is equal to five 𝑑 squared minus 𝑑.

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 𝑑π‘₯. And then we actually differentiate both of our parametric equations. And that gives us in this case 𝑑𝑦 𝑑𝑑 and then 𝑑π‘₯ 𝑑𝑑. But remembering that we actually had to find the reciprocal of 𝑑π‘₯ 𝑑𝑑 to find 𝑑𝑑 𝑑π‘₯, then finally we apply the chain rule to get our final answer.

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