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

Video Transcript

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

So what we actually have here is a pair of parametric equations. So to differentiate this to find 𝑑𝑦 𝑑𝑥, what we can actually do is use the chain rule which will allow us to differentiate them individually. And then, put it together. So to actually use the chain rule in this question, what we can say is that 𝑑𝑦 𝑑𝑥 is gonna be equal to 𝑑𝑦 𝑑𝑡 multiplied by 𝑑𝑡 𝑑𝑥. So therefore, as I said, what it does is it allows us to actually differentiate each of our parametric equations separately. And then bring them together to find 𝑑𝑦 𝑑𝑥.

So what I’m gonna do is start by differentiating each of our equations. So I’m gonna start with 𝑦 equals four 𝑡 squared plus five 𝑡. So I’m gonna differentiate this to find 𝑑𝑦 𝑑𝑡 which is gonna give us eight 𝑡 plus five. And just as a quick reminder of how we actually did that, I’m looking at the first term. So what we’ve got is the coefficient multiplied by the exponent, so a four by two, which gives us eight 𝑡. And then, what we’ve done is actually reduce the exponent by one, so two minus one. So it just gives us 𝑡 to the power of one or 𝑡. Okay, so great. We’ve now differentiated 𝑦 equals four 𝑡 squared plus five 𝑡. Now, let’s look at the next equation.

Well, the next equation is 𝑥 is equal to four 𝑡 squared plus one. So therefore, if we differentiate this, we’re gonna find 𝑑𝑥 𝑑𝑡. And this is gonna be equal to eight 𝑡. And we don’t have anything else to add on. Because if we differentiate one, we just get zero.

Okay, so we now got 𝑑𝑦 𝑑𝑡 and 𝑑𝑥 𝑑𝑡. However, if we look at the chain rule, what it says is that 𝑑𝑦 𝑑𝑥 is gonna be equal to 𝑑𝑦 𝑑𝑡 multiplied by 𝑑𝑡 𝑑𝑥. So we can actually say, “Ah, this is gonna be the reciprocal of what we found cause we found 𝑑𝑥 𝑑𝑡”. But, if we want to actually find what 𝑑𝑡 𝑑𝑥 is, what we can actually do is just find the reciprocal of 𝑑𝑥 𝑑𝑡. So therefore, we can say that 𝑑𝑡 𝑑𝑥 is equal to one over eight 𝑡. Okay, fantastic. We’ve got everything we need.

So what we can now do is actually find 𝑑𝑦 𝑑𝑥 by applying our chain rule. So therefore, we can say that 𝑑𝑦 𝑑𝑥 is equal to eight 𝑡 plus five multiplied by one over eight 𝑡. And that’s because it’s 𝑑𝑦 𝑑𝑡 multiplied by 𝑑𝑡 𝑑𝑥. So we can say finally that given that 𝑥 equals four 𝑡 squared plus one and 𝑦 equals four 𝑡 squared plus five 𝑡, then 𝑑𝑦 𝑑𝑥 is gonna be equal to eight 𝑡 plus five over eight 𝑡.