### Video Transcript

Find ππ¦ ππ₯, given that π¦ equals π₯ cubed plus seven π₯ squared plus six over π₯ plus eight.

Now, in order to actually solve this problem, what weβre gonna have to do is differentiate our function. But to enable us to do that, what weβre gonna use is the quotient rule. And we can use the quotient rule because our function is in the form π’ over π£. So as weβve got that, what weβre gonna have a look at is what the quotient rule actually is. Well, the quotient rule tells us that ππ¦ ππ₯ is equal to π£ ππ’ ππ₯ minus π’ ππ£ ππ₯ over π£ squared. So what this actually means in practice is π£ multiplied by the derivative of π’ minus π’ multiplied by the derivative of π£ all over π£ squared.

Okay, great. So now, weβve got the quotient rule. Letβs apply it to actually find ππ¦ ππ₯. So the first thing Iβve done is Iβve actually identified π’ and π£ in our question. So our π’ is gonna be π₯ cubed plus seven π₯ squared plus six because thatβs our numerator. And our π£ is gonna be our denominator which is π₯ plus eight. So next, what Iβm gonna do, Iβm actually gonna find ππ’ ππ₯. So Iβm gonna differentiate our π’ value weβve got here. So weβll differentiate π₯ cubed plus seven π₯ squared plus six.

Just a quickly recap, how weβre gonna differentiate each of these. What weβre gonna look at is the differentiations of general rules, so our power rule. And what it is, is if weβve got a function thatβs in the form ππ₯ to the power of π, then the first derivative of that function is gonna be equal to πππ₯ to the power of π minus one. So our coefficient multiplied by our exponent. And then, we reduce the exponent by one because itβs gonna be πππ₯ to the power of π minus one. Okay, weβve just recapped that. Letβs move on and differentiate and find ππ’ ππ₯.

So we go ahead and apply that power rule. And our first term is gonna be three π₯ squared. And thatβs because our coefficient is one multiplied by the exponent three gives us three. And then we reduce the exponent by one, so π₯ to the power of two or π₯ squared. And our second term is just 14π₯, so positive 14π₯. The six, when we differentiate the six, it actually disappears cause if you differentiate just an integer on its own, we get zero. Okay, great. So weβve now found ππ’ ππ₯.

So now, we move on to ππ£ ππ₯. So again, we differentiate π₯ plus eight. And when we do that, we just get one. And thatβs because π₯ differentiates to one. And as before, the positive eight just differentiates to zero. Great, so weβve now found ππ’ ππ₯ and ππ£ ππ₯. So now, we can move on to the final stage which is to actually apply the quotient rule to find out ππ¦ ππ₯.

So when we actually apply the quotient rule, weβre gonna get ππ¦ ππ₯ is equal to. And then first of all, π£ ππ’ ππ₯. So our π£, π₯ plus eight, multiplied by our ππ’ ππ₯, which is three π₯ squared plus 14π₯. And then weβre gonna subtract π₯ cubed plus seven π₯ squared plus six because thatβs our π’ ππ£ ππ₯. And itβs just that value there because our ππ£ ππ₯ is just one. Okay, great. And then finally, what we do is we actually divide by π£ squared. So itβs π₯ plus eight squared. Okay, fabulous. Weβre actually at the point now where weβve put all of our values into our quotient rule. Letβs simplify to find ππ¦ ππ₯.

Well, our first stage is to actually expand the parentheses. So weβre gonna have π₯ multiplied by three π₯ squared which is gonna give us three π₯ cubed. And then, we have π₯ multiplied by positive 14π₯ which gives us plus 14π₯ squared. And then next, we have positive eight multiplied by three π₯ squared which gives us positive 24π₯ squared. So finally, we get plus 112π₯. And thatβs because we had positive eight multiplied by positive 14π₯. Then we have minus π₯ cubed minus seven π₯ squared minus six.

So at this point, we actually can draw your attention to a possible common mistake. So as you can see, thereβs a negative in front of the parentheses. I put the second part in a parentheses because actually just to remind you that actually everything inside there has to be negative. So itβs negative π₯ cubed minus seven π₯ squared minus six. Because commonly, people will just do minus π₯ cubed. But then, theyβll just do plus seven π₯ squared plus six. So be careful for that. Okay, and then this is all divided by π₯ plus eight squared.

And when we simplify this, we have three π₯ cubed minus π₯ cubed which gives us two π₯ cubed. Then we have 14π₯ squared plus 24π₯ squared minus seven π₯ squared which gives us positive 31π₯ squared. Then we have plus 112π₯ minus six over π₯ plus eight all squared.

So therefore, we can say that given that π¦ is equal to π₯ cubed plus seven π₯ squared plus six over π₯ plus eight, then ππ¦ ππ₯ is equal to two π₯ cubed plus 31π₯ squared plus 112π₯ minus six over π₯ plus eight all squared.