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Current Year 13 thread Mark I (2014-2015)

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Original post by O133
I'll give it a go.


you're an angel!

edit: lol forgot to add the Q

y=((3x-2)^2)/sqrt x

differentiate with respect to x

I can differentiate it but cannot simplify it for the life of me!!
(edited 9 years ago)
Original post by Planckton
you're an angel!


Ask away... :smile:
Original post by O133
Ask away... :smile:


edited my post above cause I'm stupid and forgot to ask :lol:
Hi, I'm Beth. My AS results were A (Biology), A (Psychology), B (Chemistry) and B (History). I've dropped History and picked up AS Maths this year, and I'm hoping to achieve A* in both Biology and Psychology and at least an A in Chemistry. I'm getting really stressed out about personal statements because I want to apply to Oxford for Biochemistry (the B in Chemistry will probably reduce my chances, but it's worth a try anyway). I suppose the advantage of sending an early application, though, is that I won't need to stress about writing my personal statement when I've sent my application. :smile:
Original post by Planckton
you're an angel!

edit: lol forgot to add the Q

y=((3x-2)^2)/sqrt x

differentiate with respect to x

I can differentiate it but cannot simplify it for the life of me!!


The simplification I have is (sqrt x)/2[27x^2-12x-4)

I'm not sure I can spot anything simpler, and I'm posting my working (by hand because I don't know LaTeX) because I've probably slipped up somewhere.

Do you have an answer to aim for?
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Original post by O133
The simplification I have is (sqrt x)/2[27x^2-12x-4)

I'm not sure I can spot anything simpler, and I'm posting my working (by hand because I don't know LaTeX) because I've probably slipped up somewhere.

Do you have an answer to aim for?


Apparently the answer is ((9x+2)(3x-2))/2 sqrt (x^3)?
Original post by Planckton
Apparently the answer is ((9x+2)(3x-2))/2 sqrt (x^3)?


The brackets multiply to form the quadratic I had, but I really don't understand the square root bit (I had x^(1/2) and they have x^(-3/2)).
Original post by O133
The brackets multiply to form the quadratic I had, but I really don't understand the square root bit (I had x^(1/2) and they have x^(-3/2)).


Yeah! Completely understand what you've done- makes sense! Don't understand how that happens!
Original post by Planckton
you're an angel!

edit: lol forgot to add the Q

y=((3x-2)^2)/sqrt x

differentiate with respect to x

I can differentiate it but cannot simplify it for the life of me!!


An aide memoire for quotient rule: "low d high minus high d low over low squared." It's a bit of a rap, but very helpful.

If you suspect the 'answer' given in the back of your textbook or wherever is wrong, I suggest using WolframAlpha.
(edited 9 years ago)
I should really look at things before I post :mmm:
Original post by Bude8
An aide memoire for quotient rule: "low d high minus high d low over low squared." It's a bit of a rap, but very helpful.

If you suspect the 'answer' given in the back of your textbook or wherever is wrong, I suggest using WolframAlpha.


I thought I was the only one that still used that :lol:


Posted from TSR Mobile
WOOT !!! :bumps:

I just finished on UCAS: details, info, finance & 5 University choices(took quite an substantial time)
only...education, employment & personal statement to go. Yay me :bebored:

4 Down only 3 more to go !! :biggrin:

Has anyone else finished their application including Personal statement yet?
Original post by Bude8
An aide memoire for quotient rule: "low d high minus high d low over low squared." It's a bit of a rap, but very helpful.

If you suspect the 'answer' given in the back of your textbook or wherever is wrong, I suggest using WolframAlpha.


It rhymes better if you use "low d high minus high d low, over the square of what's below". Just saying. :wink:
Original post by Tibbz2
If my predicted grades are A*A*A is it worth applying to both Imperial and Manchester\? just incase Imperial hate me :tongue:


I would consider Oxford Cambridge,UCL, KCL


Posted from TSR Mobile
Original post by Planckton
you're an angel!

edit: lol forgot to add the Q

y=((3x-2)^2)/sqrt x

differentiate with respect to x

I can differentiate it but cannot simplify it for the life of me!!



Original post by O133
The simplification I have is (sqrt x)/2[27x^2-12x-4)

I'm not sure I can spot anything simpler, and I'm posting my working (by hand because I don't know LaTeX) because I've probably slipped up somewhere.

Do you have an answer to aim for?


I wouldn't bother using the quotient rule tbh - just write it as x^(-1/2)(2x-2)^2 and use the product rule.
Original post by usycool1
I wouldn't bother using the quotient rule tbh - just write it as x^(-1/2)(2x-2)^2 and use the product rule.


You so clever :lol: . I've forgotten 100% my A-level maths :lol:
Original post by drowninginsound
It rhymes better if you use "low d high minus high d low, over the square of what's below". Just saying. :wink:


I trust your musical abilities, but I'm too used to it and saying 'over the square of what's below' just sounds like an extra mouthful :lol:
Original post by Nirgilis
You so clever :lol: . I've forgotten 100% my A-level maths :lol:


Meh, my A-Level Maths/Further Maths knowledge has gone SO rusty in the last few months. :tongue:
Good afternoon
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Original post by foreverstrong_97
Good afternoon


I assume you had a good day? :h:

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