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1. Hi, I'm stuck with several things.

1. v = ln (2x-3) dv/dx=2/2x-3

I thought ln x = 1/x so ln 2x-3 = 1/2x-3... Why?? Please somebody explain this.

If you could give me some clues on how to tackle 2 & 3, would help so much.

2. Given that lxl <1, expand [(1+x)/(1-x)]^(1/3) in ascending powers of x and up to and including the term in x^2.

3. The coefficients of x and x^2 in the expansion of (1+px+qx^2)^-2 in ascending pwoers of x are 4 and 14 respectively. Find the values of p and q.

Thank you!!!
2. (Original post by TheQueen1986)
Hi, I'm stuck with several things.

1. v = ln (2x-3) dv/dx=2/2x-3

I thought ln x = 1/x so ln 2x-3 = 1/2x-3... Why?? Please somebody explain this.

Thank you!!!
Chain rule means when you differentiate...you leave the differential of whats in the brackets on top of the fraction...and divide it by f(x)

I think..it's been awhile since i did maths

G
3. (Original post by TheQueen1986)
Hi, I'm stuck with several things.

1. v = ln (2x-3) dv/dx=2/2x-3

I thought ln x = 1/x so ln 2x-3 = 1/2x-3... Why?? Please somebody explain this.

If you could give me some clues on how to tackle 2 & 3, would help so much.

2. Given that lxl <1, expand [(1+x)/(1-x)]^(1/3) in ascending powers of x and up to and including the term in x^2.

3. The coefficients of x and x^2 in the expansion of (1+px+qx^2)^-2 in ascending pwoers of x are 4 and 14 respectively. Find the values of p and q.

Thank you!!!
ln x is the natural logarithm of x. If you are having problem with a basic function like the logarithm function then you need to go and revise quite a bit.

Nevertheless if y = ln x then dy/dx = 1/x, that may be where your confusion came from. So if y = ln (2x-3) then we apply the chain rule i.e. y = ln t, t=2x-3
then dy/dx = 1/t * dt/dx and dt/dx = 2 so dy/dx = 2/t = 2/(2x-3)

2. Clue: Taylor series, take the first few derivatives and evaluate at 0.

3. Again you need to use Taylor series, take a few derivatives treating p and q as constants and it should follow.

If you need more help just ask or pm me.
4. (Original post by AntiMagicMan)
ln x is the natural logarithm of x. If you are having problem with a basic function like the logarithm function then you need to go and revise quite a bit.

Nevertheless if y = ln x then dy/dx = 1/x, that may be where your confusion came from. So if y = ln (2x-3) then we apply the chain rule i.e. y = ln t, t=2x-3
then dy/dx = 1/t * dt/dx and dt/dx = 2 so dy/dx = 2/t = 2/(2x-3)

2. Clue: Taylor series, take the first few derivatives and evaluate at 0.

3. Again you need to use Taylor series, take a few derivatives treating p and q as constants and it should follow.

If you need more help just ask or pm me.

taylor series is not taught in P3 only binomial expansion can be used for this
5. It isn't... meh they must have changed the syllabus since I did it.
6. OK thanks, I get about the logarithm stuff.

But for 2 and 3, I don't know the Taylor series, never been taught that. Does anyone know a binomial way of approaching?? Thanks.
7. (Original post by TheQueen1986)
OK thanks, I get about the logarithm stuff.

But for 2 and 3, I don't know the Taylor series, never been taught that. Does anyone know a binomial way of approaching?? Thanks.
For number 2- split the part in the brackets into partial fractions by long division of the term...this should give you something like "1 + 1/x+1" ....something like that...i haven't done it so that's probably wrong..but i'd imagine it was in that form.

Then you can use binomial expansion to figure it out..i think

G
8. (Original post by TheQueen1986)
Hi, I'm stuck with several things.

1. v = ln (2x-3) dv/dx=2/2x-3

I thought ln x = 1/x so ln 2x-3 = 1/2x-3... Why?? Please somebody explain this.

If you could give me some clues on how to tackle 2 & 3, would help so much.

2. Given that lxl <1, expand [(1+x)/(1-x)]^(1/3) in ascending powers of x and up to and including the term in x^2.

3. The coefficients of x and x^2 in the expansion of (1+px+qx^2)^-2 in ascending pwoers of x are 4 and 14 respectively. Find the values of p and q.

Thank you!!!
For 2. Binomial expansion can bedone by converting into
[(1+x)^(1/3)][(1-x)^(-1/3)]

Then expand the two separately and multiply them.
If you need help on the actual expansion then PM me.
9. Q3:
(1+ px+qx^2)^-2.
Put px + qx^2 = t. Using binomial expansion:
(1 + t)^-2 = 1 + (-2)t + (-2)(-3)t^2/2! + ...
= 1 + (-2)(px+qx^2) + 3(px +qx^2)^2 + ...
= 1 + (-2p)x + (-2q + 3p^2)x^2 +...

So you got the power of x and x^2. Replace then solve.
Try this.
10. Sorry, still have some probs.

Q2:

I did what you did, but I still have troubles.

Ok I expand the numerator and the denominator. But to what x^??

Do you mean when you multiply the two, that you multiply numerator by 1/denominator? So if the expanded denominator was something like (just making it up for now) 1+2x+3x^2 you make it 1+2x^-1+3x^-2.

Q3:

How did you get from:= 1 + (-2)(px+qx^2) + 3(px +qx^2)^2 + ...
----> 1 + (-2p)x + (-2q + 3p^2)x^2 +...
11. (Original post by TheQueen1986)
2. Given that lxl <1, expand [(1+x)/(1-x)]^(1/3) in ascending powers of x and up to and including the term in x^2.
2.

[(1+x)/(1-x)]^(1/3) = [(1+x)^(1/3)][(1-x)^(-1/3)]

Now using binomial expansiona dn slittping the two expressions seperately:

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

(1-x)^(-1/3) = 1 + x/3 + [(-1/3)(-4/3)x^2]/2 + ...

Now multiplying the two expanded expressions and collecting the terms only upto x^2

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

= 1 +x/3 + (2/9)(x^2) + x/3 + (1/9)(x^2) + (1/9)(x^2)

= 1 + (2/3)x + (4/9)(x^2)

[(1+x)/(1-x)]^(1/3) = 1 + (2/3)x + (4/9)(x^2) + ...

================================ ==========================
12. Does the above post makes sense to u TheQueen1986?
13. (Original post by TheQueen1986)
Sorry, still have some probs.

Q2:

I did what you did, but I still have troubles.

Ok I expand the numerator and the denominator. But to what x^??

Do you mean when you multiply the two, that you multiply numerator by 1/denominator? So if the expanded denominator was something like (just making it up for now) 1+2x+3x^2 you make it 1+2x^-1+3x^-2.

Q3:

How did you get from:= 1 + (-2)(px+qx^2) + 3(px +qx^2)^2 + ...
----> 1 + (-2p)x + (-2q + 3p^2)x^2 +...
Hic, it's the formula for binomial expansion in P2 but u must have condition:
|px + qx^2| < 1. But in ur question, i didnt find that, maybe there's another way for this (not Taylor series).
But if it uses binomial expansion, as u know
(1 + x)^n = 1 + nx + n(n-1)x^2/2! + n(n-1)(n-2)x^3/3! + .....
But i think it's right, cuz it's P2 question.
14. Yeah I got Q2 and Q3 sorted completely now. Thanks guys.
15. A fellow A level student trying to finish their A2 maths before the year starts? Heh.

Q1: Use the chain rule, don't make false assumptions:
V = ln(2x-3)
V = ln(t), t=2x-3
dv/dt = 1/t, dt/dx =2
dv/dx = 2/t,
dv/dx = 2/(2x-3)

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