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junior college 2 | H1 Maths
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QN
QN

junior college 2 chevron_right H1 Maths chevron_right Singapore

Need help thx

Date Posted: 3 years ago
Views: 512
J
J
3 years ago
10x - 6 = A(5x - 2) + B

10x - 6 = 5Ax - 2A + B

Comparing coefficients,

10 = 5A
A = 2


-6 = -2A + B
B = 2A - 6
= 2(2) - 6
= 4 - 6
= -2
QN
QN
3 years ago
Wow thx
QN
QN
3 years ago
So this has nothing to do with the first equation above?
J
J
3 years ago
So 10x - 6 = 2(5x - 2) - 2


y = (10x - 6) / (5x - 2)²

= (2(5x - 2) - 2) / (5x - 2)²

= 2/(5x - 2) - 2/(5x - 2)²

= 2(5x - 2)-¹ - 2(5x - 2)-²



dy/dx = 2(-1)(5x - 2)-²(5) - 2(-2)(5x - 2)-³(5)

= -10/(5x - 2)² + 20/(5x - 2)³

= 20/(5x - 2)³ - 10/(5x - 2)²

= (20 - 10(5x - 2)) / (5x - 2)³

= (40 - 50x) / (5x - 2)³
QN
QN
3 years ago
For part 2, can i
d/dx[(10x-6)(5x-2)^-2]

= [-2(10x-6)(5x-2)^-3](5)

= -10(10x-6)(5x-2)^-3

= -(100x-60)(5x-2)^-3

= - (100x-60)/(5x-2)^3
J
J
3 years ago
No. You forgot to differentiate the (10x - 6). You have to apply product rule.

But the idea for this question is, part i) already allows you to rewrite the expression as partial fractions.

So we don't have to do product rule or quotient rule. Just differentiate the two terms separately

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Ignatius Yeo
Ignatius Yeo's answer
36 answers (Tutor Details)
1st
Hope it helps!
QN
QN
3 years ago
How did you get 2(5x-2)-2 instead of 2(5x-3)
Ignatius Yeo
Ignatius Yeo
3 years ago
You must fit into the form(A(5x-2)+B).
Ignatius Yeo
Ignatius Yeo
3 years ago
2(5x-2)-2 is also the same as 2(5x-3). However, the question is asking for 2(5x-2)-2.
QN
QN
3 years ago
Ahh I see
QN
QN
3 years ago
Thanks!
Ignatius Yeo
Ignatius Yeo
3 years ago
Because the idea here is to simplify a more complex expression into simpler identities, to make it easy for you to differentiate. Same applies when doing integration too.
Ignatius Yeo
Ignatius Yeo
3 years ago
And the identities can be fitted into different forms for your own use. For eg, making it easier for you to differentiate. Or fitting into the exact form when doing integration.( Like in the form of f'(x)/f(x) for integration.)
QN
QN
3 years ago
Thanks!
A side qn: how do I differentiate when two unknowns are multiplied to each other?

Let’s say d/dx[ (10x)(5x-2)^-2 ]

Do I ignore the 10x and

= -2(10x)(5x-2)^-3(5)

= -100x(5x-2)^-3

= - 100x/ (5x-2)^3
Ignatius Yeo
Ignatius Yeo
3 years ago
For differentiation, or even Integration you do not ignore any terms. Either you make it into some forms similar to the general formula or by expanding the terms and differentiate one by one by following the rules of differentiation. For the question you have given, you can use quotient rule but will be tedious. You can change 10x into 2(5x-2)+4 in order to perform differentiation.
Ignatius Yeo
Ignatius Yeo
3 years ago
Usually if you notice, the form you change to involves the denominator, so that you can further simplify into other identities to allow you to perform differentiation easily. For eg (nx+c)/(ax+b)^j= (Z(ax+b)+D/(ax+b)^j). Note: ax+b/ax+b=1
QN
QN
3 years ago
I see!
But what if it cannot be changed to another form that is the same as the denominator?
Do I have to use the quotient rule like the one I sent?
Ignatius Yeo
Ignatius Yeo
3 years ago
It is very unlikely it cannot be changed to other form that involves the denominator. Even if it involves surds, it still can be done but the numbers may not look nice and you may not get used to it.
QN
QN
3 years ago
Thank a lot for the help really appreciate it
Ignatius Yeo
Ignatius Yeo
3 years ago
No problem:) anyway good luck for your studies and hope that you have fully understood.
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QN
Qn's answer
19 answers (A Helpful Person)
By using the quotient rule, did you mean like this?
Ignatius Yeo
Ignatius Yeo
3 years ago
Yes this was what I meant, your quotient rule is correct. However, the second last step you could have factorise (5x-2) so the final answer is (-50x-20)/((5x-2)^3).
Ignatius Yeo
Ignatius Yeo
3 years ago
Anyway if you have time, recheck the answer for the quotient rule. How do you arrive at 10-100x/(5x-2)^3?