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