Chain, Product and Quotient Rule: Exercise 4

00:00 Let's have a look at our next example. We have y equals to 4x to the minus 2, 7 plus 3x to the minus 5. Observe and look at your function for a bit and decide what kind of function you see and what kind of method you would use now to differentiate it. Now you may think or you may have a look at it and see that there are actually a few different ways of writing it and then you have a choice of using different rules. For example, I could take this x to the minus 2 down. So I could write this as 4 over x squared and then 7 plus 3x to the minus 5. Leave that x to the minus 5 there because it's in the brackets just to not overcomplicate things. So this is the same as 4, 7 plus 3x to the minus 5 over x squared. You can see that both of these are the same functions. So this function here is exactly the same as this function, only rewritten. And both of them will now need different methods for differentiation. So, this function here is a function multiplied by another function.

01:09 So u times v. However, this function is a function that's being divided by another.

01:15 Both of them would give you the same answer if you use the product rule here and if you use the quotient rule here. You should always just double check using the alternate method, M, just to use that you've got the same answer. Now, I personally feel that the product rule is always easier because it doesn't have to divide by the v squared. So I'm going to go with the product rule and if there are ways in which you can convert your quotient rule question into a product rule, you always have the freedom to do that. Like I said, you know lots of different methods now that you can apply as and when you please. So, if I stick to the product rule, I say u equals to 4x to the minus 2 and v equals to 7 plus 3x to the minus 5. Differentiate each one of them separately. So you bring the power down, that gives me minus 8x to the minus 3. Remember to take 1 away from the power. This in fact is a little bit misleading because it looks like a chain rule, but there is in fact no power on the outside. So this is simply 7 plus 3x to the minus 5. So it's not really chain rule, there is no function of a function. It's just a normal function here that's not within another function. So when you differentiate this, the 7 disappears because it's just a constant. Bring the 5 down, so that gives you minus 15x to the minus 6. Okay, and now we apply the product rule. So we're going to do vdudx and udvdx. So we can now say dy by dx is minus 8x to the minus 3 multiplied by 7 plus 3x to the minus 5. You then add 4x to the minus 2 and we're multiplying this with minus 15x to the minus 6. Okay, all we have to do is tidy this up. So minus 8x to the minus 3, going to leave this as it is. You have 4 multiplied by 15 which gives you 60, but remember it's going to be minus 60 because of this minus number there. And you can now also multiply these 2 together. So that's going to be x to the minus 2 multiplied by x to the minus 6 which gives you x to the minus 8. You can just leave it here because you've got lots of negative powers here, you could take them down so you could write it as minus 8 over x cubed and you have 7 plus 3 over x to the 5 and then you've got minus 60 over x to the 8. So, we're almost done. You can expand this out by multiplying this through if you would like to. If we multiply minus 8 with 7, we get minus 56 over x cubed. We have minus 8 times 3. So you multiply the top to give you minus 24 but also remember to multiply this exes together. This gives you x to the power of 8 and then you have minus 60 over x to the power of 8. So, once we're at this point, we can do one last step if you observe that we have x to the 8x to the 8 as a common denominator. So you can join those final 2 terms together. If we do that here, we get minus 56 over x cubed and then you have minus 24 minus 60 to give you minus 84 over x to the 8. And that is our simplified version of this answer. So, this was us solving this problem using the product rule. You should get the same answer when you use the quotient rule. It may be a variation of it so simplify differently but numerically it should be the same value.