Lectures

Chain, Product and Quotient Rule: Exercise 4

by Batool Akmal
(1)

Questions about the lecture
My Notes
  • Required.
Save Cancel
    Learning Material 2
    • PDF
      DLM Quotient Rule, Chain Rule and Product Rule Exercise Calculus Akmal.pdf
    • PDF
      Download Lecture Overview
    Report mistake
    Transcript

    00:01 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.

    00:21 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 choice of using different rules.

    00:29 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. Let?s leave the x to the minus 5 there because it?s in the brackets just to not over complicate things.

    00:45 So this is the same as 4, 7 plus 3x to the minus 5, over x squared.

    00:51 You can see that both of these are the same functions.

    00:56 So this function here is exactly the same as this function, only rewritten and both of them will now need different methods for differentiation.

    01:04 So this function here is a function multiplied by another function, so u times v.

    01:10 However this function is a function that's being divided by another.

    01:14 Both of them would give you the same answer if you use the product rule here and if you use the quotient rule here.

    01:22 We should always just double check using the alternate method just to see that you?ve got the same answer.

    01:29 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 gonna 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.

    01:44 Like I said, you know lots of different methods now that you can apply as and when you please.

    01:49 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.

    01:59 Differentiate each one of them separately so you bring the power down.

    02:04 That gives me minus 8x to the minus 3. Remember to take one away from the power.

    02:09 This, in fact is a little bit misleading because it looks like a chain rule but there's in fact no power on the outside. So this is simply 7 plus 3x to the minus 5.

    02:20 So it?s not really chain rule, there is no function of a function.

    02:23 It?s just a normal function here, that?s not within another function.

    02:27 So when you differentiate this, the 7 disappears ?cause it?s just a constant.

    02:32 Bring the 5 down, so that gives you minus 15x to the minus 6.

    02:37 Okay, and now we apply the product rule. So we're going to do vdudx and udvdx, so we can now say dy/dx is minus 8x to the minus 3, multiplied by 7 plus 3x to the minus 5.

    02:55 You then add 4x to the minus 2 and we're multiplying this with minus 15x to the minus 6.

    03:07 Okay, all we have to do is tidy this up, so minus 8x to the minus 3, going to leave this as it is.

    03:14 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 two 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.

    03:38 You can just leave it here because you?ve got lots of negative powers here.

    03:43 You could take them down so you could write it as minus 8 over x cubed, and then you have 7 plus 3 over x to the 5, and then you've got minus 60 over x to the 8.

    03:56 So we're almost done. You can expand this out by multiplying these through, if you would like to.

    04:02 If we multiply minus 8 with 7 we get minus 56 over x cubed.

    04:08 We have minus 8 time 3, so you multiply the top to give you minus 24 but also remember to multiply these x's together. This gives you x to the power of 8 and then you have minus 60 over x to the power of 8.

    04:22 So once we are at this point we can do one last step if you observe that we have x to the 8x, x to the 8 as a common denominator, so you can join those final two 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.

    04:48 So this was us solving this problem using the product rule.

    04:52 You should get the same answer when you use the quotient rule.

    04:55 It may be a variation of it, so simplify differently but numerically it should be the same value.


    About the Lecture

    The lecture Chain, Product and Quotient Rule: Exercise 4 by Batool Akmal is from the course Quotient Rule, Chain Rule and Product Rule.


    Author of lecture Chain, Product and Quotient Rule: Exercise 4

     Batool Akmal

    Batool Akmal


    Customer reviews

    (1)
    5,0 of 5 stars
    5 Stars
    5
    4 Stars
    0
    3 Stars
    0
    2 Stars
    0
    1  Star
    0