### Video Transcript

Differentiate ππ₯ equals four π₯
squared minus five π₯ plus eight over three π₯ minus four.

So in order to actually
differentiate our function, what weβre gonna use is something called the quotient
rule. 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.

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. 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. Okay, fab! Now that we have this, letβs use it
and actually find out what the derivative of our function actually is.

So my first step is to actually
identify our π’ and our π£. So I use our numerator, so four π₯
squared minus five π₯ plus eight. And our π£ is going to be three π₯
minus four, which is our denominator. So the first thing weβre gonna do
is actually differentiate π’. So we differentiate four π₯ squared
minus five π₯ plus eight.

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. So you subtract one from the
exponent.

Okay, so now letβs use this and
differentiate π’. So weβre gonna get eight π₯ minus
five. 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. So two minus one is one. So weβre left with eight π₯.

Okay, great! So now letβs move on to ππ£
ππ₯. Well, if we differentiate π£, so
weβre gonna find ππ£ ππ₯. And this is just gonna be equal to
three. And thatβs because if we
differentiate three π₯, we get three. And if we differentiate just an
integer on its own, you get zero.

Okay, so great! Weβve now found ππ’ ππ₯ and ππ£
ππ₯. So we can now actually apply our
quotient rule to find ππ¦ ππ₯. So first of all, weβre gonna have
π£ ππ’ ππ₯. So itβs gonna be three π₯ minus
four multiplied by eight π₯ minus five. And then weβre gonna have minus
three multiplied by four π₯ squared minus five π₯ plus eight. So thatβs our π’ ππ£ ππ₯. And Iβve put it the other way round
just because itβs easier when weβre gonna expand the parentheses in the next
stage. And then this is all divided by
three π₯ minus four all squared. So itβs our π£ squared.

Okay, so now letβs move on to the
next stage, which Iβve already said is gonna be expanding the parentheses on the
numerator. So first of all, Iβve got three π₯
multiplied by eight π₯, which will give me 24π₯ squared. 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. And a negative multiplied by
negative gives us positive.

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. Okay, then this is all divided by
three π₯ minus four all squared.

So we now are gonna move on to our
final stage, which is actually to simplify our numerator. So then weβre gonna have our first
term. Itβs gonna be 12π₯ squared. And thatβs because weβve got 24π₯
squared minus 12π₯ squared. Then we get minus 32π₯. And thatβs because we had negative
15π₯ minus 32π₯ plus 15π₯. So therefore, the 15π₯s cancel
out. We have minus 32π₯.

And then, finally, we get minus
four. And this is because we had positive
20 minus 24. And then this is all over three π₯
minus four all squared. So great! 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.