Edexcel C4 - Exercise 4B - Question 1F (page 40 of the new textbook)

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Stevey-R
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#1
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#1
This question has confused me for the past hour or so, I simply can't match my answer with the back of the textbook. Any help would be appreciated

Find and expression in terms of x and y for dy/dx, given that:

x = 2y / (x^2 - y)

Thanks

Also: the answer given in the back is (3x^2 - y) / (2+x)
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ttoby
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#2
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#2
The way I would approach it is to times both sides by (x^2-y), multiply out all brackets, then differentiate with respect to x.

Then put all dy/dx terms on one side, factorise to get it in the form (dy/dx)(......)=(.....) then you can get a formula for dy/dx.
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Stevey-R
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#3
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(Original post by ttoby)
The way I would approach it is to times both sides by (x^2-y), multiply out all brackets, then differentiate with respect to x.

Then put all dy/dx terms on one side, factorise to get it in the form (dy/dx)(......)=(.....) then you can get a formula for dy/dx.
Thanks tried your method and got the right answer much appreciated. I was struggling away with the quotient rule and just just overwhelmed with the complexity of it all :P
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-suze-
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#4
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Hi,

I understand how to get the answer based on the above method, however when I first attempted this question I tried to use the quotient rule (and then the product rule) but couldn't get the answer in the book.

I realise the above method is easier than using the quotient or product rules, I'm just trying to work out where I'm going wrong.

Here is my working:

Image

Many thanks for any help.
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ttoby
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#5
Report 7 years ago
#5
(Original post by Stevey-R)
This question has confused me for the past hour or so, I simply can't match my answer with the back of the textbook. Any help would be appreciated

Find and expression in terms of x and y for dy/dx, given that:

x = 2y / (x^2 - y)

Thanks

Also: the answer given in the back is (3x^2 - y) / (2+x)
(Original post by -suze-)
Hi,

I understand how to get the answer based on the above method, however when I first attempted this question I tried to use the quotient rule (and then the product rule) but couldn't get the answer in the book.

I realise the above method is easier than using the quotient or product rules, I'm just trying to work out where I'm going wrong.

Here is my working:

Image

Many thanks for any help.
I read through your workings and couldn't find a mistake.

It's possible for this sort of question to get more than one correct answer.

Since we know that x = 2y / (x^2 - y) always then we could, for instance, take any correct answer and replace x with 2y / (x^2 - y) and get another correct answer.

I haven't tried using this sort of manipulation to show that the textbook answer and your answer are equivilant, but (since this wasn't a 'show that' question) that's unnecessary as you've been able to justify your answer by showing your lines of correct workings.

As it's a question in a textbook then they're probably just including the answer they expect you to get. However if something like this came up in an exam then the mark scheme would list a few alternate methods and answers and tell the examiner to be aware of this issue.
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ghostwalker
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#6
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#6
(Original post by -suze-)
Here is my working:
Just to confirm what ttoby said. I have worked it through and your answer can, as it must, be manipulated to get the same as theirs. Though it's actually easier to convert theirs to yours.

Outline:

You're told x=...

Hence x+2=...

Now sub that into their answer - that sorts the denominator out.

For the numerator now split  (3x^2-y)(x^2-y) into 2x^2(x^2-y)+(x^2-y)(x^2-y)

For the first set of terms, note from the given equation (rearranged), we have (x^2-y)x=2y. Sub in, and you're done.
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-suze-
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#7
Report 7 years ago
#7
Brilliant, thank you very much, I can see now that my answer and the book answer are the same.

I did try to substitute some values of x and y into my answer and the book answer and got different results, so had assumed my answer was wrong, but I've realised I made a mistake when substituting the x and y values, so it all makes sense now.

Thanks!
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