WEBVTT
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Differentiate ππ₯ equals four π₯ squared minus five π₯ plus eight over three π₯ minus four.
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So in order to actually differentiate our function, what weβre gonna use is something called the quotient rule.
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And we know that weβre gonna use the quotient rule because actually our function is in the form π’ over π£, because itβs actually a fraction.
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What the quotient rule actually tells us is that ππ¦ ππ₯, so our derivative, is equal to π£ ππ’ ππ₯ minus π’ ππ£ ππ₯ over π£ squared.
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What that actually means in practice is its our π£ multiplied by our derivative of π’ and then itβs minus π’ multiplied by our derivative of π£ and then all divided by π£ squared.
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Okay, fab!
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Now that we have this, letβs use it and actually find out what the derivative of our function actually is.
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So my first step is to actually identify our π’ and our π£.
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So I use our numerator, so four π₯ squared minus five π₯ plus eight.
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And our π£ is going to be three π₯ minus four, which is our denominator.
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So the first thing weβre gonna do is actually differentiate π’.
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So we differentiate four π₯ squared minus five π₯ plus eight.
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And just to quickly remind us how we do that, if weβve got ππ₯ in the form ππ₯ to the power of π, then the derivative of our function, or could be known as ππ¦ ππ₯, is equal to ππ, so coefficient multiplied by the exponent, and then π₯ to the power of π minus one.
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So you subtract one from the exponent.
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Okay, so now letβs use this and differentiate π’.
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So weβre gonna get eight π₯ minus five.
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And just to kind of remind us how we did that, the first one was gonna be eight π₯, because weβve got four multiplied by two, an exponent, multiplied by coefficient, which gives us eight, and then π₯ to the power of one, because you actually subtract one from the exponent.
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So two minus one is one.
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So weβre left with eight π₯.
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Okay, great!
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So now letβs move on to ππ£ ππ₯.
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Well, if we differentiate π£, so weβre gonna find ππ£ ππ₯.
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And this is just gonna be equal to three.
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And thatβs because if we differentiate three π₯, we get three.
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And if we differentiate just an integer on its own, you get zero.
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Okay, so great!
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Weβve now found ππ’ ππ₯ and ππ£ ππ₯.
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So we can now actually apply our quotient rule to find ππ¦ ππ₯.
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So first of all, weβre gonna have π£ ππ’ ππ₯.
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So itβs gonna be three π₯ minus four multiplied by eight π₯ minus five.
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And then weβre gonna have minus three multiplied by four π₯ squared minus five π₯ plus eight.
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So thatβs our π’ ππ£ ππ₯.
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And Iβve put it the other way round just because itβs easier when weβre gonna expand the parentheses in the next stage.
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And then this is all divided by three π₯ minus four all squared.
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So itβs our π£ squared.
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Okay, so now letβs move on to the next stage, which Iβve already said is gonna be expanding the parentheses on the numerator.
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So first of all, Iβve got three π₯ multiplied by eight π₯, which will give me 24π₯ squared.
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Then three π₯ multiplied by negative five gives us negative 15π₯, then minus 32π₯ because we had negative four multiplied by eight π₯, and then finally plus 20 because we had negative four multiplied by negative five.
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And a negative multiplied by negative gives us positive.
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Then weβve got minus 12π₯ squared, because weβve got negative three multiplied by four π₯ squared, plus 15π₯, negative three multiplied by negative five π₯, again negative multiplied by negative, and then finally minus 24.
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Okay, then this is all divided by three π₯ minus four all squared.
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So we now are gonna move on to our final stage, which is actually to simplify our numerator.
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So then weβre gonna have our first term.
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Itβs gonna be 12π₯ squared.
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And thatβs because weβve got 24π₯ squared minus 12π₯ squared.
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Then we get minus 32π₯.
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And thatβs because we had negative 15π₯ minus 32π₯ plus 15π₯.
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So therefore, the 15π₯s cancel out.
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We have minus 32π₯.
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And then, finally, we get minus four.
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And this is because we had positive 20 minus 24.
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And then this is all over three π₯ minus four all squared.
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So great!
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We can say that if we differentiate the function four π₯ squared minus five π₯ plus eight over three π₯ minus four, then ππ¦ ππ₯ is gonna be equal to 12π₯ squared minus 32π₯ minus four over three π₯ minus four all squared.