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# Year 13 Maths Help Thread

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1. (Original post by SeanFM)
The product rule is when you have two functions multiplying eachother - you're mixing it up with the chain rule a bit, though you do have to (spoiler alert) use the chain rule here.

If you say f(x) = (x+1)^3 and g(x) = (x+2)^6.. you do the rest
I think thats what i did?

I made u = x+1 then v = u^3
then using the product rule i got the equation from my other post. Unless you make u = f(x) and v = g(x)?
2. (Original post by kiiten)
I think thats what i did?

I made u = x+1 then v = u^3
then using the product rule i got the equation from my other post. Unless you make u = f(x) and v = g(x)?
I would suggest revisitng the product rule

What you have mentioned in your post there is the chain rule, which is used in this question but you have to apply to product rule first.

If you are ever stuck with something, check the definitions of whatever you're using then look for simple examples and understand them, then look for similar examples.
3. (Original post by SeanFM)
I would suggest revisitng the product rule

What you have mentioned in your post there is the chain rule, which is used in this question but you have to apply to product rule first.
Yeah I agree

Is the product rule uv^1 + vu^1 ?
If so then i made u = x+1 and v = u^3
4. (Original post by kiiten)
Yeah I agree

Is the product rule uv^1 + vu^1 ?
If so then i made u = x+1 and v = u^3
I think you know what you mean but you're not typing / expressing it correctly. If u and v are functions, (uv)' = u'v + v'u is the product rule.

But with that second line you are defining the chain rule.. those are two different things. Again, I would suggest looking up the two to refresh your memory
5. (Original post by SeanFM)
I think you know what you mean but you're not typing / expressing it correctly. If u and v are functions, (uv)' = u'v + v'u is the product rule.

But with that second line you are defining the chain rule.. those are two different things. Again, I would suggest looking up the two to refresh your memory
Ohh so u = f(x) and v = g(x)? yeah i know - please could you check if i have the right idea before i check it up (i dont want to confuse myself haha)

you find u' and v' by using the chain rule
then apply the product rule
6. (Original post by kiiten)
Ohh so u = f(x) and v = g(x)? yeah i know - please could you check if i have the right idea before i check it up (i dont want to confuse myself haha)

you find u' and v' by using the chain rule
then apply the product rule
You can do it in either orderer. But yes, those two lines of what you said are correct, so I think you do understand. Although I think it is more logical to apply the product rule first.

So as long as you understand why those two things are true you can carry on and do the working.
7. ok so im resitting s1 and c1 this year because i bottled them :/ when do you reckon i should start revising/learning them again? now?
8. (Original post by SeanFM)
You can do it in either orderer. But yes, those two lines of what you said are correct, so I think you do understand. Although I think it is more logical to apply the product rule first.

So as long as you understand why those two things are true you can carry on and do the working.
How would you do it the other way round (with product rule first)? - i mean how would you find u' or v' without the chain rule??
9. (Original post by kiiten)
How would you do it the other way round (with product rule first)? - i mean how would you find u' or v' without the chain rule??
I'm not sure if we're on the same page then.

The way I picture it, the problem here is like building drawers for a... wardrobe. You can build the frame of the wardrobe first or build the shelves but you will need to put together both at some point.

The product rule just tells you that to differentiate (x+1)^3 * (x+2)^6 (defining u = (x+1)^3 and v = (x+2)^6) you need to find uv' + vu'. There you have applied the product rule and now you do not need to use it again. But now you need to find what v' and u' are before multiplying them by u and v respectively. This is why you can try to find u' and v' before applying the product rule to (x+1)^3 * (x+2)^6 but why it makes less logical sense than to apply the product rule first.
10. Could someone help me prove by induction that sum from n to 2n of n^2 equals [n/6](14n+1)(n+1)

I got to the point (14k^3+39k^2+24k+6)/6 and some have non real solutions so I'm guessing it's wrong so I'm not sure where I went wrong.

For the induction part I tried to simplify [k/6](14k+1)(k+1) + (2k+1)^2
11. (Original post by metrize)
Could someone help me prove by induction that sum from n to 2n of n^2 equals [n/6](14n+1)(n+1)

I got to the point (14k^3+39k^2+24k+6)/6 and some have non real solutions so I'm guessing it's wrong so I'm not sure where I went wrong.

For the induction part I tried to simplify [k/6](14k+1)(k+1) + (2k+1)^2
I assume you mean ?

If so then your induction part isnt quite correct.

For n=k you are assuming

and for n = k+1 you are doing , so what terms do you need to add and subtract from your n = k summation
12. (Original post by not_lucas1)
ok so im resitting s1 and c1 this year because i bottled them :/ when do you reckon i should start revising/learning them again? now?
I would say finish C3 then look at C1 again. After you're confident with C1 (it honestly won't take long..... you just need to remember some specifics, you'll have the ability to do so instead of trying to grow it) start on S1, this should be maybe Jan-Feb time. That's how I'd approach it anyway
13. (Original post by DylanJ42)
I assume you mean ?

If so then your induction part isnt quite correct.

For n=k you are assuming

and for n = k+1 you are doing , so what terms do you need to add and subtract from your n = k summation
Ah I see. Thanks a lot I didn't notice that
14. (Original post by SeanFM)
I'm not sure if we're on the same page then.

The way I picture it, the problem here is like building drawers for a... wardrobe. You can build the frame of the wardrobe first or build the shelves but you will need to put together both at some point.

The product rule just tells you that to differentiate (x+1)^3 * (x+2)^6 (defining u = (x+1)^3 and v = (x+2)^6) you need to find uv' + vu'. There you have applied the product rule and now you do not need to use it again. But now you need to find what v' and u' are before multiplying them by u and v respectively. This is why you can try to find u' and v' before applying the product rule to (x+1)^3 * (x+2)^6 but why it makes less logical sense than to apply the product rule first.
Like this??? - ive found u and u' etc. and just need to sub them into the product rule.

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15. (Original post by kiiten)
Like this??? - ive found u and u' etc. and just need to sub them into the product rule.

Posted from TSR Mobile
It looks like you've calculated the same thing twice using different notation..but yes, now you should have all you need.
16. (Original post by SeanFM)
It looks like you've calculated the same thing twice using different notation..but yes, now you should have all you need.
I thought so. For example the first part (uv') i end up with (x+1)^3 6 (x+2)^5

D: helpp

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17. (Original post by kiiten)
I thought so. For example the first part (uv' i end up with (x+1)^3 6 (x+2)^5

D: helpp

Posted from TSR Mobile
I'm not sure what you want me to help you with or how
18. (Original post by SeanFM)
I'm not sure what you want me to help you with or how
Sorry i must have forgot to add something. Umm please could you tell me the answer and explain how you get there? Im really confused, I looked over the product rule but this example confuses me
19. (Original post by kiiten)
Sorry i must have forgot to add something. Umm please could you tell me the answer and explain how you get there? Im really confused, I looked over the product rule but this example confuses me
You already have all the tools available.

You know that you are looking for uv' + vu'. You know what u, u', v and v' are.

In your previous post you found uv' correctly. Now you just need vu'.

The only thing that you may or may not have seen before about this example is how you use the chain rule and the product rule in the same question. But you just need to apply the product rule to uv to give uv' + vu' and then you need to use the chain rule when finding v' and u'.
20. (Original post by SeanFM)
You already have all the tools available.

You know that you are looking for uv' + vu'. You know what u, u', v and v' are.

In your previous post you found uv' correctly. Now you just need vu'.

The only thing that you may or may not have seen before about this example is how you use the chain rule and the product rule in the same question. But you just need to apply the product rule to uv to give uv' + vu' and then you need to use the chain rule when finding v' and u'.
(x+1)^3 6 (x+2)^5 + (x+2)^6 3(x+1)^2

But thats not in the form (x+1)^2 (x+2)^5 (ax + b) D:

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