differentiation c3 helppp

JSM1

im getting the wrong answer. Can someone tell me what im doing wrong?

differentation .jpg133.9KB

Reply 1

Pangol

Original post by JSM1

im getting the wrong answer. Can someone tell me what im doing wrong?

The error is in your fourth line, when you factorise out (2x + 1)^(-2).

Reply 2

aislingrafter

I've just started this topic in school too not sure if I'm right did it without a calculator lol

Original post by JSM1

im getting the wrong answer. Can someone tell me what im doing wrong?

Reply 3

JSM1

Original post by Pangol

The error is in your fourth line, when you factorise out (2x + 1)^(-2).

what have i done wrong? I used the exact method in previous questions where you expand

Reply 4

Pangol

Original post by JSM1

what have i done wrong? I used the exact method in previous questions where you expand

You are factorising something that has the form A x^(-2) + B x^(-1). If you factorised x^(-2) out of this, what would you have?

(It's not a huge error, but one missing symbol has lead you to head off in the wrong direction.)

Reply 5

JSM1

Original post by Pangol

do you add the terms?

help

Original post by JSM1

do you add the terms?

Sorry - thought I had replied to this. I think you are right, but what do you mean by "the terms"? They change their form after factorising.

Reply 8

JSM1

Original post by Pangol

Sorry - thought I had replied to this. I think you are right, but what do you mean by "the terms"? They change their form after factorising.

do you add over here

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Reply 9

Pangol

Original post by JSM1

do you add over here

Yeah, that's the bit that is wrong. You should be adding the (-2x + 2) and the (2x + 1), not multiplying them - but even if you were supposed to be multiplying them (which you are not!), there should be a bracket around the first part.

Reply 10

JSM1

Original post by Pangol

do you know the reason why you add these terms as in previous questions i expanded the bracket by the number in front of it.

Reply 11

Pangol

Original post by JSM1

do you know the reason why you add these terms as in previous questions i expanded the bracket by the number in front of it.

I can't be sure without seeing the question, but I expect you got the right answer by coincidence.

This is just factorisation. In the example I mentioned above, A x^(-2) + B x^(-1), we can factorise out x^(-2) from both terms. How many of these do we have? In the first term, we have A of them. In the second term, we have Bx of them. So altogether, we have A + Bx of them, that is, A x^(-2) + B x^(-1) = x^(-2) [A + Bx].

Reply 12

JSM1

Original post by Pangol

Question:

Find the value of dy/dx at the point x=2 on the curve with the equation

y=(x-1)(2x+1)^-1

Reply 13

Pangol

Original post by JSM1

Question:

Find the value of dy/dx at the point x=2 on the curve with the equation

y=(x-1)(2x+1)^-1

Yeah, well to check this I'd need to do the whole thing! I'd rather see your working. But more to the point - do you get what I am saying about factorisation?

Reply 14

JSM1

Original post by Pangol

Yeah, well to check this I'd need to do the whole thing! I'd rather see your working. But more to the point - do you get what I am saying about factorisation?

yes i do understand abit but previous questions have the same format so why do you multiply for those

Reply 15

JSM1

this is the other question.

differen.jpg188.2KB

Reply 16

JSM1

over here i have multiplied out 1(2x+1) which gave me 2x+1,

then adding 1x.. 2x+1+1x gave me........... 3x+1

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help?

Original post by JSM1

over here i have multiplied out 1(2x+1) which gave me 2x+1,

then adding 1x.. 2x+1+1x gave me........... 3x+1

This is right. When you factorise out the 3(2x + 1)^(-1/2), you are left with x from the first term, and 2x + 1 from the second term. You then correctly add these together.

In the problem question, you are left with -2x + 2 from the first term, and 2x + 1 from the second term. You should be adding these together. What you have done is a strange mixture of taking the -2x from the first term, and multiplying the second term by the 2 from the first term.