WEBVTT
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Determine ππ¦ ππ₯ given that π¦ equals seven multiplied by log to base four of seven π₯ minus three.
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Well, the first thing weβre gonna do when we actually want to differentiate our function here is actually change the base.
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And weβre gonna change the base from log to base four to ln.
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So the base is gonna be π.
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And we know that actually when we change the base, if we have log to the base π of π₯, then this is gonna be equal to ln π₯ over ln π.
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And therefore, weβre gonna get π¦ is equal to seven ln seven π₯ plus three over ln four.
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And thatβs because our π₯ was seven π₯ minus three and our π was four.
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Okay, great!
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So now actually letβs look at how we differentiate this.
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So what we can say when we differentiate it is that ππ¦ ππ₯ is gonna be equal to β and what Iβve done is taken out the constant because weβve got seven over ln four, which is just gonna be a value, then multiply this by d ππ₯ of ln seven π₯ plus three.
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So what we need to do now is actually differentiate ln seven π₯ plus three.
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And to enable us to do that, we actually have a couple of relationships that we know.
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So we know that if π¦ equals ln π₯, then ππ¦ ππ₯ is equal to one over π₯.
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And we also know that if π¦ equals ln π of π₯, then ππ¦ ππ₯ is gonna be equal to the derivative of π of π₯ over π of π₯.
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And we actually get this from the first one using the chain rule.
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Now weβre actually gonna use the second relationship because ours is actually in the form ln π of π₯ cause weβve got ln seven π₯ plus three.
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So therefore, we can say that ππ¦ ππ₯ is gonna be equal to seven multiplied by seven over seven π₯ plus three multiplied by ln four.
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And we got seven multiplied by seven on our numerator because we had seven in our constant, and this is multiplied by the derivative of our function seven π₯ plus three.
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Well if you differentiate seven π₯ plus three, youβre just left with seven because if we look at seven π₯, well seven multiplied by one, so the coefficient multiplied by the exponent just gives us a seven.
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And if you reduce the exponent by one, so you go from one to zero, then weβre just left with seven as three differentiates to zero.
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Okay, great!
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Thatβs our numerator.
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And then we got our denominator as our π of π₯ was seven π₯ plus three.
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Okay, great!
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So what weβve now got is ππ¦ ππ₯.
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But now what we need to do is actually simplify it.
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So therefore, we can say that given that π¦ equals seven log to the base four of seven π₯ minus three, then ππ¦ ππ₯, so the derivative, is equal to 49 over seven π₯ plus three ln four.
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So just a quick recap of how we did that.
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So first of all, we change the base.
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So if you have something thatβs log to the base four et cetra, then change the base.
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So we change the base so that we actually had the base of π.
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So we use the natural logarithm.
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Then we took the constant out, which was seven over natural log of four, just because this was actually easier when we actually moved on to the differentiation of the other part.
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So then we actually differentiated ln seven π₯ plus three.
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And we did that using a relationship we know.
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However, like I said, the relation we know couldβve been found using the chain rule.
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And we couldβve done that using the chain rule because if we have a function of the form π¦ equals π of π’, then ππ¦ ππ₯ is equal to ππ¦ dπ’ multiplied by dπ’ ππ₯.
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So in our case, if weβre gonna have ππ¦ dπ’, well that wouldβve just been the derivative of ln π’, which would just be one over π’.
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And then dπ’ ππ₯ would be the derivative of seven π₯ plus three, which-which is what we had as our numerator.
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Hence, we got the derivative of our function over the function itself.
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Okay, great!
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And then our final step was just to simplify to leave us with an answer: ππ¦ ππ₯ is equal to 49 over seven π₯ plus three ln four.