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junior college 2 | H1 Maths
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Need help thx
Date Posted: 4 years ago
Views: 565
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
10x - 6 = 5Ax - 2A + B
Comparing coefficients,
10 = 5A
A = 2
-6 = -2A + B
B = 2A - 6
= 2(2) - 6
= 4 - 6
= -2
Wow thx
So this has nothing to do with the first equation above?
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
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
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
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
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
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
See 2 Answers
done
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Hope it helps!
Date Posted:
4 years ago
How did you get 2(5x-2)-2 instead of 2(5x-3)
You must fit into the form(A(5x-2)+B).
2(5x-2)-2 is also the same as 2(5x-3). However, the question is asking for 2(5x-2)-2.
Ahh I see
Thanks!
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.
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.)
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
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
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.
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
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?
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?
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.
Thank a lot for the help really appreciate it
No problem:) anyway good luck for your studies and hope that you have fully understood.
done
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By using the quotient rule, did you mean like this?
Date Posted:
4 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).
Anyway if you have time, recheck the answer for the quotient rule. How do you arrive at 10-100x/(5x-2)^3?
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