Video Transcript
Find ππ¦ ππ₯ if π¦ is equal to four to the power of nine π₯ squared minus four π₯ plus eight.
Now, in order to actually find ππ¦ ππ₯, what weβre gonna do is actually weβre gonna use the chain rule. And the chain rule states that ππ¦ ππ₯ is equal to ππ¦ ππ’ multiplied by ππ’ ππ₯. And this is when π¦ is equal to a function of π’ and π’ is equal to a function of π₯. Okay, so we now got the chain rule. Letβs apply it to our problem to actually solve and find ππ¦ ππ₯.
So first of all, we need to identify our π’. So thatβs gonna be nine π₯ squared minus four π₯ plus eight. And therefore, our π¦ is gonna be equal to four to the power of π’. Okay, thatβs the first step. Now, what we need to do is actually differentiate to find ππ¦ ππ’ and ππ’ ππ₯.
Iβm gonna start by finding ππ’ ππ₯ and I will do that by differentiating the expression nine π₯ squared minus four π₯ plus eight. And this is just gonna give us 18π₯ minus four. Weβve done that the normal way. So we just differentiated or highlighted that with the first term. So what we did is we multiplied the exponent by the coefficient. So two multiplied by nine is 18 and then we reduced the exponent by one β so from two to one. So we just get 18π₯.
So great, weβve differentiated that. Now, letβs move on and find ππ¦ ππ’. So now to find ππ¦ ππ’, what weβre gonna have to do is actually differentiate four to the power of π’. And in order to do this, weβre actually gonna use the general rule which is that if weβre gonna differentiate π to the power of π₯, then weβre gonna get an answer of π to the power of π₯ multiplied by the natural logarithm of π. So therefore, as our result, weβre gonna get four to power of π’ because thatβs our π and then multiply it by the natural logarithm of four because again thatβs our π because itβs π to the power of π₯ ln π.
Great, so weβve now differentiated both parts. We can use the chain rule to put them together to find ππ¦ ππ₯. So now, weβre actually gonna apply the chain rule, which tells us that ππ¦ ππ₯ is equal to ππ¦ ππ’ multiplied by ππ’ ππ₯. So weβre gonna get ππ¦ ππ₯ is equal to four to the power of π’ ln four and thatβs because that was our ππ¦ ππ’ and then multiplied by 18π₯ minus four because that was our ππ’ ππ₯.
Okay, great, but this isnβt a final answer because we can see there are actually still including π’. So what we now need to do is actually substitute in our value for π’. So therefore, weβre gonna get β and Iβve just rearranged a bit here β 18π₯ minus four multiplied by four to the power of nine π₯ squared minus four π₯ plus eight because that was our π’ and then thatβs multiplied by the natural logarithm of four.
So therefore, after some tidying up using factoring, we can say that if π¦ is equal to four to the power of nine π₯ squared minus four π₯ plus eight, therefore ππ¦ ππ₯ is gonna be equal to two multiplied by nine π₯ minus two. And we got that because actually we factored 18π₯ minus four because two is a factor of 18π₯ and negative four. And this is multiplied by four to the power of nine π₯ squared minus four π₯ plus eight ln four.