# Differentiation and Factorization Problem

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#1
It's the thanksgiving holiday and I am still in my house studying for maths...
We started differentiation about two weeks ago, and are currently doing further differentiation (the chain, power, and quotient rule, as well as the implicit thingy). I do not have any problems on how to use those formulas, but have problems in factorizing the final answers (when using the power and quotient formulas). When factorizing, I seem to be getting the wrong answers according to the math textbook. So since I know that the forum is my friend, I'm here to ask you what did I do wrong.

Q1.

1. y = x^3(3-x^4)^5

2. dy/dx = x^3(5)(3-x^4)^4(-4x^3) + (3-x^4)^5(3x^2)

3. = -20x^3(3-x^4)^4 + (3x^2)(3-x^4)^5

4. = (x^2)(3-x^4)^4(-20x + 3(3-x^4))

5. = (x^2)(3-x^4)^4(-20x + 9 - 3x^4))

Q2.

1. y = (x^2-7)(x^4+1)^2

2. dy/dx = 8x^3(x^2-7)(x^4+1) + 2x(x^4+1)^2

3. = x^2 - 7 + 2x(x^4 + 1)(4x^2(x^4+1))
0
17 years ago
#2
(Original post by FurbyMaX)
It's the thanksgiving holiday and I am still in my house studying for maths...
We started differentiation about two weeks ago, and are currently doing further differentiation (the chain, power, and quotient rule, as well with the implicit thingy). I do not have any problems on how to use those formulas, but have problems in factorizing the final answers (when using the power and quotient formulas). When factorizing, I seem to be getting the wrong answers according to the math textbook. So since I know that the forum is my friend, I'm here to ask you what did I do wrong.

Q1.

1. y = x^3(3-x^4)^5

2. dy/dx = x^3(5)(3-x^4)^4(-4x^3) + (3-x^4)^5(3x^2)

3. = -20x^3(3-x^4)^4 + (3x^2)(3-x^4)^5

4. = (x^2)(3-x^4)^4(-20x + 3(3-x^4))

5. = (x^2)(3-x^4)^4(-20x + 9 - 3x^4))

Q2.

1. y = (x^2-7)(x^4+1)^2

2. dy/dx = 8x^3(x^2-7)(x^4+1) + 2x(x^4+1)^2

3. = x^2 - 7 + 2x(x^4 + 1)(4x^2(x^4+1))

Are you joking? You started differentiation 2 weks ago and now yu are on this??
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#3
(Original post by *dave*)
Are you joking? You started differentiation 2 weks ago and now yu are on this??
No, Im not joking, why?
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#4
(Original post by FurbyMaX)
No, Im not joking, why?
...?
0
17 years ago
#5
God!

After 2 weeks i was still trying to work out the equation for the tangent to the curve lol ... surely you need A2 knowledge of the binomical expansion in order to factorise some of those expressions, you surely wouldnt do them the long way would you?
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#6
(Original post by *dave*)
God!

After 2 weeks i was still trying to work out the equation for the tangent to the curve lol ... surely you need A2 knowledge of the binomical expansion in order to factorise some of those expressions, you surely wouldnt do them the long way would you?
Im doing the Ib, what's A2? I've got the terms messed up, am I doing factorization the long way? Thanks
0
17 years ago
#7
(Original post by FurbyMaX)
Im doing the Ib, what's A2? I've got the terms messed up, am I doing factorization the long way? Thanks
To multiply out this bracket:

y = x^3(3-x^4)^5

Surely you would need knowledge of binomical exapnsion, surely you would write out (3-x^4)(3-x^4)(3-x^4)(3-x^4)(3-x^4)

??
0
17 years ago
#8
(Original post by *dave*)
To multiply out this bracket:

y = x^3(3-x^4)^5

Surely you would need knowledge of binomical exapnsion, surely you would write out (3-x^4)(3-x^4)(3-x^4)(3-x^4)(3-x^4)

??
Or you could use t = (3-x^4) god im being thick today lol
0
17 years ago
#9
(Original post by *dave*)
To multiply out this bracket:

y = x^3(3-x^4)^5

Surely you would need knowledge of binomical exapnsion, surely you would write out (3-x^4)(3-x^4)(3-x^4)(3-x^4)(3-x^4)

??
Why do you want to multiply that thingy out anyway? Isn't that already fully facterised? Oh and it's Bionomial not bionomical.

(Original post by *dave*)
Are you joking? You started differentiation 2 weks ago and now yu are on this??
You can pretty much finish P1 in a month and P2 (where chain, product and quotient rules are) in another month. No big deal. A'Level these days are just too easy. Maths is getting a little bit stupid I think...Ever seen anything from 20 years ago? Gosh...they were hard.
0
#10
(Original post by *dave*)
To multiply out this bracket:

y = x^3(3-x^4)^5

Surely you would need knowledge of binomical exapnsion, surely you would write out (3-x^4)(3-x^4)(3-x^4)(3-x^4)(3-x^4)

??
I guess, yes. Is there another way?
0
17 years ago
#11
y = x^3(3-x^4)^5

u = x^3
du/dx = 3x^2

v = (3-x^4)^5
w = 3-x^4
dw/dx = -4x^3
v = w^5
dv/dw = 5w^4

dv/dx = dw/dx . dv/dw
=(-4x^3)(5w^4)
=(-4x^3)(5(3-x^4)^4)

then product rule
dy/dx = v du/dx + u dv/dx
=(3-x^4)^5(3x^2) + x^3(-4x^3)(5(3-x^4)^4)

that should work
0
17 years ago
#12
(Original post by FurbyMaX)
It's the thanksgiving holiday and I am still in my house studying for maths...
We started differentiation about two weeks ago, and are currently doing further differentiation (the chain, power, and quotient rule, as well as the implicit thingy). I do not have any problems on how to use those formulas, but have problems in factorizing the final answers (when using the power and quotient formulas). When factorizing, I seem to be getting the wrong answers according to the math textbook. So since I know that the forum is my friend, I'm here to ask you what did I do wrong.

Q1.

1. y = x^3(3-x^4)^5

2. dy/dx = x^3(5)(3-x^4)^4(-4x^3) + (3-x^4)^5(3x^2)

3. = -20x^3(3-x^4)^4 + (3x^2)(3-x^4)^5

4. = (x^2)(3-x^4)^4(-20x + 3(3-x^4))

5. = (x^2)(3-x^4)^4(-20x + 9 - 3x^4))

Q2.

1. y = (x^2-7)(x^4+1)^2

2. dy/dx = 8x^3(x^2-7)(x^4+1) + 2x(x^4+1)^2

3. = x^2 - 7 + 2x(x^4 + 1)(4x^2(x^4+1))
Q1:
You made a mistake in step 3. The first term should be -(20x^6)(3-x^4)^4 (you forgot to square x^3). So you get:-
dy/dx = -(20x^6)(3-x^4)^4 + (3x^2)(3-x^4)^5
= [(3-x^4)^4][(3x^2)(3-x^4) - (20x^6)]

Q2:
I don't understand how you went from step 2 to 3. You should get:-
dy/dx = (8x^3)(x^2-7)(x^4+1) + 2x(x^4+1)^2
= (x^4+1)[(8x^3)(x^2-7) + 2x(x^4+1)]

It would be advisable to do binomial expansions only if the question explicitly demands it. I agree that A-level maths today is too easy, which is why the top universities require additional exams.
0
17 years ago
#13
For Q.1 I get:

dy/dx = x^2(3-x^4)^5(3-20x^4(3-x^4))

Q2. dy/dx = 2x(x^4+1)(4x^2(x^2-7)+x^4+1)
0
#14
Thanks guys, I'll see if the answers u gave me are correct tomorrow, now I just wanna play some games 0
17 years ago
#15
(Original post by Camford)
Why do you want to multiply that thingy out anyway? Isn't that already fully facterised? Oh and it's Bionomial not bionomical.

You can pretty much finish P1 in a month and P2 (where chain, product and quotient rules are) in another month. No big deal. A'Level these days are just too easy. Maths is getting a little bit stupid I think...Ever seen anything from 20 years ago? Gosh...they were hard.

I found my A level maths hard... 0
17 years ago
#16
I found my A level maths hard... Not the end of everything though, is it?
0
17 years ago
#17
(Original post by Camford)
Not the end of everything though, is it?

Well, no... it's do-able, if that's what you mean! And even if you fail it, it isn't the end of the world... but I wouldn't call it easy. Having said that, I'm not that mathematically strong (ask my poor patient maths supervisor)!
0
17 years ago
#18
However easy some people find it, and I'm certain a small minority of people find it very easy, it has to be said Maths is one of the tougher A-Levels with lots to learn, and combined with a poor national standard of teaching, makes Maths one of the toughest A-Levels, especially given the two jumps (from gcse - as and as-a2) are both reasonably big.
0
17 years ago
#19
i make the answer to q1
dy/dx = [(3x^2)(3-x^4)^5] - [(20x^6)(3-x^4)^4]
or factorised a bit (pointlessly):

dy/dx = [x^2(3-x^4)^4] * [3(3-x^4) - 20x^4]
0
17 years ago
#20
(Original post by FurbyMaX)
It's the thanksgiving holiday and I am still in my house studying for maths...
We started differentiation about two weeks ago, and are currently doing further differentiation (the chain, power, and quotient rule, as well as the implicit thingy). I do not have any problems on how to use those formulas, but have problems in factorizing the final answers (when using the power and quotient formulas). When factorizing, I seem to be getting the wrong answers according to the math textbook. So since I know that the forum is my friend, I'm here to ask you what did I do wrong.

Q1.

1. y = x^3(3-x^4)^5

2. dy/dx = x^3(5)(3-x^4)^4(-4x^3) + (3-x^4)^5(3x^2)

3. = -20x^3(3-x^4)^4 + (3x^2)(3-x^4)^5

4. = (x^2)(3-x^4)^4(-20x + 3(3-x^4))

5. = (x^2)(3-x^4)^4(-20x + 9 - 3x^4))

Q2.

1. y = (x^2-7)(x^4+1)^2

2. dy/dx = 8x^3(x^2-7)(x^4+1) + 2x(x^4+1)^2

3. = x^2 - 7 + 2x(x^4 + 1)(4x^2(x^4+1))
=1
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