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