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

Question Video: Finding the First Derivative of a Function Defined by Parametric Equations Mathematics • Third Year of Secondary School

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Given that π₯ = 4π‘Β² + 1 and π¦ = 4π‘Β² + 5π‘, find ππ¦/ππ₯.

02:35

Video Transcript

Given that π₯ equals four π‘ squared plus one and π¦ equals four π‘ squared plus five π‘, find ππ¦ ππ₯.

So what we actually have here is a pair of parametric equations. So to differentiate this to find ππ¦ ππ₯, 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 ππ¦ ππ₯ is gonna be equal to ππ¦ ππ‘ multiplied by ππ‘ ππ₯. 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 ππ¦ ππ₯.

So what Iβm gonna do is start by differentiating each of our equations. So Iβm gonna start with π¦ equals four π‘ squared plus five π‘. So Iβm gonna differentiate this to find ππ¦ ππ‘ which is gonna give us eight π‘ plus five. And just as a quick reminder of how we actually did that, Iβm looking at the first term. So what weβve got is the coefficient multiplied by the exponent, so a four by two, which gives us eight π‘. And then, what weβve done is actually reduce the exponent by one, so two minus one. So it just gives us π‘ to the power of one or π‘. Okay, so great. Weβve now differentiated π¦ equals four π‘ squared plus five π‘. Now, letβs look at the next equation.

Well, the next equation is π₯ is equal to four π‘ squared plus one. So therefore, if we differentiate this, weβre gonna find ππ₯ ππ‘. And this is gonna be equal to eight π‘. And we donβt have anything else to add on. Because if we differentiate one, we just get zero.

Okay, so we now got ππ¦ ππ‘ and ππ₯ ππ‘. However, if we look at the chain rule, what it says is that ππ¦ ππ₯ is gonna be equal to ππ¦ ππ‘ multiplied by ππ‘ ππ₯. So we can actually say, βAh, this is gonna be the reciprocal of what we found cause we found ππ₯ ππ‘β. But, if we want to actually find what ππ‘ ππ₯ is, what we can actually do is just find the reciprocal of ππ₯ ππ‘. So therefore, we can say that ππ‘ ππ₯ is equal to one over eight π‘. Okay, fantastic. Weβve got everything we need.

So what we can now do is actually find ππ¦ ππ₯ by applying our chain rule. So therefore, we can say that ππ¦ ππ₯ is equal to eight π‘ plus five multiplied by one over eight π‘. And thatβs because itβs ππ¦ ππ‘ multiplied by ππ‘ ππ₯. So we can say finally that given that π₯ equals four π‘ squared plus one and π¦ equals four π‘ squared plus five π‘, then ππ¦ ππ₯ is gonna be equal to eight π‘ plus five over eight π‘.

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