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# Maths P1 Algebra (syllabus example) watch

1. Edexcel expect you to differentiate expressions such as:

(x^2 + 5x -3) / 3x^(1/2).

To be able to do this it needs to be organised into something easier. Is this correct?

=> [(x^2 + 5x -3) / 1] / [3x^(1/2) / 1].

=> [(x^2 + 5x -3) / 1] * [1 / 3x^(1/2)]. (flipped the second one over)

=> (x^2 + 5x -3) * 3x^(-1/2)

= 3x^(3/2) + 15x^(1/2) - 9x^(-1/2).

Is this correct? It is essential to master expressions such as these for the exam.
2. It looks ok but not normal P1 stuff. This sort of question doesn't normally come up until p3 when you have learnt the quotient rule.

MB
3. => [(x^2 + 5x -3) / 1] * [1 / 3x^(1/2)]
=> (x^2 + 5x -3) * 3x^(-1/2)

this part is wrong I think.
4. Can you not rationalise (sp?) the denominator in this expression and then differentiate it?
5. [QUOTE=Whywhywhy]

=> [(x^2 + 5x -3) / 1] * [1 / 3x^(1/2)]. (flipped the second one over)

=> (x^2 + 5x -3) * 3x^(-1/2)
QUOTE]

in the second line, you forgot to change 3 to 3^-1. your method is correct, but i would split it straight away into

(x^2 + 5x -3)/3x^0.5 = x^1.5/3 + (5/3)x^-0.5 - x^0.5

and differentiate this.
6. (Original post by Leekey)
Can you not rationalise (sp?) the denominator in this expression and then differentiate it?
yeah, that's what i mean!
7. (Original post by Leekey)
Can you not rationalise (sp?) the denominator in this expression and then differentiate it?
you could...but you'd still end up with '3x' at the bottom anyway, so in the end you're basically simplifying each term first, and then differentiating (which could have been done without rationalising)
8. (Original post by mockel)
you could...but you'd still end up with '3x' at the bottom anyway, so in the end you're basically simplifying each term first, and then differentiating (which could have been done without rationalising)
Yeah but my basic instinct tells me that surds in denominators are bdad things and must be destroyed at all costs!!!
9. (Original post by Whywhywhy)
Edexcel expect you to differentiate expressions such as:

(x^2 + 5x -3) / 3x^(1/2).

To be able to do this it needs to be organised into something easier. Is this correct?

=> [(x^2 + 5x -3) / 1] / [3x^(1/2) / 1].

=> [(x^2 + 5x -3) / 1] * [1 / 3x^(1/2)]. (flipped the second one over)

=> (x^2 + 5x -3) * 3x^(-1/2)

= 3x^(3/2) + 15x^(1/2) - 9x^(-1/2).

Is this correct? It is essential to master expressions such as these for the exam.
Just divide each term by 3x^(1/2) so x^2 becomes ⅓x^(3/2), 5x becomes (5/3)x^(1/2) and -3 becomes -x^(1/2)

so you're differentiating ⅓x^(3/2)+(5/3)x^(1/2)-x^(1/2) which is easy enough.

x^0.5 / 2 + 5/6 * x^(-0.5) + x/2 ^ (-3/2) ??????
11. I was using the rule that

(a/b) / (x/y) = (a/b) * (y/x)

And I have not differentiated it yet, just rearranged.
12. I don't have the answer but I'd sure as hell like to know it.
13. (Original post by Whywhywhy)
I was using the rule that

(a/b) / (x/y) = (a/b) * (y/x)

And I have not differentiated it yet, just rearranged.
but:
[(x^2 + 5x -3) / 1] * [1 / 3x^(1/2)] =/= (x^2 + 5x -3) * 3x^(-1/2)

Here you're effectively saying a/1 * 1/b = ab, but it actually just equals a/b, as you started with in the question.
14. (Original post by Whywhywhy)
I don't have the answer but I'd sure as hell like to know it.
Like meepmeep said, use this:

(a+b+c)/d = a/d + b/d + c/d

Which is basically dividing out (like the division version of multiplying out brackets).
15. (Original post by buzfvar_1)

x^0.5 / 2 + 5/6 * x^(-0.5) + x/2 ^ (-3/2) ??????
Yes, that's correct (but be careful with the last term to make clear that it's (x^(-3/2))/2, not (x/2)^(-3/2)).
16. What I am essentially getting at is that I do not want any denominators, most certainly not anything containing x. I have no clue whatsoever about how to divide x^2 by 3x^1/2. How do I do that? I tried to get out of that by multiplying by the reciprocal...
17. (Original post by Whywhywhy)
What I am essentially getting at is that I do not want any denominators, most certainly not anything containing x. I have no clue whatsoever about how to divide x^2 by 3x^1/2. How do I do that? I tried to get out of that by multiplying by the reciprocal...
x²/√3x = (x²)(x^-½)/3 <--- I assume your question meant 3(x^½) and not (3x)^½

You can bring the denominator containing x to the top by multiplying the power of x by -1, so instead of dividing by a you'd be multiplying by the reciprocal of a. b/a = b*(a^-1)

But what you did was the wrong thing, you started with an expression with a numerator and denominator, and ended up multiplying them which is wrong. You just multiply every part of the numerator by the reciprocal of the denimonator.

From there you can just multiply out:

x²/√3x = (x²)(x^-½)/3 = (1/3)x^(2-½) = (1/3)x^(2/3)

Remember if the number is the same then you can add powers:

(a^b)(a^c) = a^(b+c)

Also bear in mind:

[(a^b)^c] = a^(bc)
18. (Original post by musicboy)
It looks ok but not normal P1 stuff. This sort of question doesn't normally come up until p3 when you have learnt the quotient rule.
yeah u are right, but to the original post, get some1 or a teacher to go through the quotient rule, as it is quite simple. even so, you shouldnt really get such hard questions.. i think the hardest dy/dx you get asked is y=x^3 /4x or something quite simple
19. (Original post by KerChing)
yeah u are right, but to the original post, get some1 or a teacher to go through the quotient rule, as it is quite simple. even so, you shouldnt really get such hard questions.. i think the hardest dy/dx you get asked is y=x^3 /4x or something quite simple
As I said this is on the SYLLABUS as an example of what has to be done. For P1.

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