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Year 13 Maths Help Thread

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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 :borat:


Ohh so u = f(x) and v = g(x)? yeah i know :tongue: - 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
Original post by kiiten
Ohh so u = f(x) and v = g(x)? yeah i know :tongue: - 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.
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?
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?? :s-smilie:
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?? :s-smilie:


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.
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
(edited 7 years ago)
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 ∑n2nr2 \displaystyle \sum_n^{2n} r^2 ?

If so then your induction part isnt quite correct.

For n=k you are assuming ∑k2kr2=k(14k+1)(k+1)6 \displaystyle \sum_k^{2k} r^2 = \frac{k(14k+1)(k+1)}{6}

and for n = k+1 you are doing ∑k+12k+2r2 \displaystyle \sum_{k+1}^{2k+2} r^2 , so what terms do you need to add and subtract from your n = k summation
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 :tongue:
Original post by DylanJ42
I assume you mean ∑n2nr2 \displaystyle \sum_n^{2n} r^2 ?

If so then your induction part isnt quite correct.

For n=k you are assuming ∑k2kr2=k(14k+1)(k+1)6 \displaystyle \sum_k^{2k} r^2 = \frac{k(14k+1)(k+1)}{6}

and for n = k+1 you are doing ∑k+12k+2r2 \displaystyle \sum_{k+1}^{2k+2} r^2 , 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
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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1475650703037.jpg39.9KB
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.
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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(edited 7 years ago)
Original post by kiiten
I thought so. For example the first part (uv':wink: 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 :s-smilie:
Original post by SeanFM
I'm not sure what you want me to help you with or how :s-smilie:


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 :s-smilie:
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 :s-smilie:


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'.
(edited 7 years ago)
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:
Original post by kiiten
(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:


No offence intended, I think you are giving up too quickly or being too quick to ask questions. Alternatively, you've hit a mental block, which is also fine, happens to the best of us. Spending too long on one problem can be frustrating and draining but equally a good learning experience.

Have a think about what you have just asked and how you can answer it for yourself :h:
(edited 7 years ago)
Original post by SeanFM
No offence intended, I think you are giving up too quickly or being too quick to ask questions. Alternatively, you've hit a mental block, which is also fine, happens to the best of us. Spending too long on one problem can be frustrating and draining but equally a good learning experience.

Have a think about what you have just asked and how you can answer it for yourself :h:


Maybe but i cant see it, I only see 2 terms that are almost in the correct form. Im out of ideas with this question :frown:, I dont know what to do anymore :argh:
Original post by kiiten
Maybe but i cant see it, I only see 2 terms that are almost in the correct form. Im out of ideas with this question :frown:, I dont know what to do anymore :argh:


Your answer in your previous post is correct, you are actually required to make it look like that form.

Here's some advice for you then, call (x+2) = a, (x+1) = b and rewrite everything in terms of a and b, from your 'almost correct form' stage and including the final form. Then see if it hits you :smile: (though maybe c and d are better to use since a and b are in the final form)
(edited 7 years ago)
Original post by SeanFM
Your answer in your previous post is correct, you are actually required to make it look like that form.

Here's some advice for you then, call (x+2) = a, (x+1) = b and rewrite everything in terms of a and b, from your 'almost correct form' stage and including the final form. Then see if it hits you :smile: (though maybe c and d are better to use since a and b are in the final form)


9x + 12 ??

I still feel like this is wrong and if i had a similar question i wouldnt know how to get it into the correct form. I understood what you said but it 'didnt hit me' (i wouldnt have thought og answering it like this) :biggrin:

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