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 vdudx minus udvdx 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...
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