And our last example, if we look at this question,
you can clearly see that there are 2 functions that are dividing, one on the other.
So, if we say y equals to 4x squared, over 10x, minus 1 cube.
Now, remember from the previous lecture, you are able to convert this into a product rule,
you can't do that with all functions but with some functions you can.
So what I mean by that, is you can rewrite this as 4x squared
and 10x minus 1 to the minus 3. On this question here you'd use the quotient rule,
because you can see that they're dividing. And on this question here,
if you were to differentiate it, you would use the product rule.
And it really is up to you, what you choose to do.
In this instance I'll go with the quotient rule,
just so that we have a little bit more practice,
but most of the times you'll find that I prefer working with the product rule,
just because it's a little bit easier. But just so we get enough practice,
we'll use the quotient rule and we'll stick with this form of this question here.
So, a reminder of quotient rule, we got dy/dx is vdu/dx minus udv/dx over v squared.
You might find with practice that you stop writing these rules,
but I find the good practice that you write it on the side,
just so that you remember, and the avoid any element of making mistakes.
So the top value is u that is 4x squared, the bottom value is v, 10x minus 1 cubed.
If you differentiate each one of them separately, that gives you 8x
and when we differentiate this, this observe closely is the chain rule,
so you've got a big function and a little function.
So, you bring the power down, you got 10x minus 1 to the 2
and then you multiply with the differential of the inside, which is 10 in our case.
Tidy this up a little bit, that give you 30, 10x minus 1 squared.
Let's put this all into our definition for the quotient rule.
So we can now say that dy/dx is vdu/dx so that multiplied by this, and udv/dx.
So, we get 8x, multiplied by 10x minus 1 cubed, multiplied by 4x squared
with 30, 10x minus 1 squared. Remember that you also have to divide this with v squared.
So we have 10x minus 1 cubed, and then you square all of them.
Almost there, we can get away with not having to expand this.
So we have 8x plus 10x minus 1 cubed. We've got 4 times 30, so we have minus a 120 squared,
and then 10x minus 1 all squared, and at the bottom, we end up with 10x minus 1.
Remember when you have a power raised by another power, you multiply them.
So we get to the power of 6. There is one little thing you could do here.
You will see that you have 10x minus 1 in each term here, here and here at the bottom.
You can simplify them out, because you've got six of them as at the bottom,
you've got 2 here and you've got 3 there.
So, you can simplify them out with the lowest factor of 10x minus 1 to the 2.
So, I mean this goes to 10x minus 1 to the power of 1. We have a 120x squared.
This entire thing just cancels out, because there's two of them,
so we can just get rid of that here.
And then at the bottom, you have 10x minus 1 to the power of 4,
because we've cancelled two brackets of 10x minus 1 with every single term.
You can see, you can do a little bit more here, you can simplify,
we can times this through because you'll have end up with an x squared term.
So that gives you 80x squared minus 8x, minus a 120x squared.
We just squeeze that in and I've got 10x minus 1 to the power of 4.
And so, you can see that these can be simplified a little bit further,
we can add this x squared terms.
So, let's just go up here for our final answer,
because we're running out of space. I have 80x squared minus a 120x squared,
which gives me minus 40x squared minus 8x and then we have our denominator,
which is 10x minus 1 to the power of 4.
And that using the quotient rule is the gradient of 4x squared
divided by 10x minus 1 cubed.
You can try using the product rule, and it should give you the same answer
or some variation of it, unsimplified version of it,
but numerically, it should be the same value.