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implicit differentiation watch

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1. Hi
i understand why they simplfied the equation and then differentiated each term with respect to x. But, i did the started by differentiating x with respect to x to get 1. And then on the left hand side used the qoutient rule to differentiate that fraction term with respect to x. it didnt reduce to the same answer.

Am i doing the quotient rule wrong or can i not do it this way.

Question:
http://pmt.physicsandmathstutor.com/...hapter%204.pdf

excercise b question 1f

2. Nothing intrinsically wrong with doing it your way. I would expect you to end up with a fairly complex expressions , however and then simplifying it will NOT be fun.
3. (Original post by DFranklin)
Nothing intrinsically wrong with doing it your way. I would expect you to end up with a fairly complex expressions , however and then simplifying it will NOT be fun.
Oh ok, it should simplify to the same expression though shouldn't it? i cant get that to happen.
4. (Original post by 111davey1)
Oh ok, it should simplify to the same expression though shouldn't it? i cant get that to happen.
If you differentiated the quotient properly with respect to x and with respect to y, it should
5. (Original post by h3rmit)
If you differentiated the quotient properly with respect to x and with respect to y, it should
I think i did.. but clearly i didnt
6. (Original post by 111davey1)
I think i did.. but clearly i didnt
7. (Original post by h3rmit)
i dont have a way of uploading it . But i think the problem lies with my quotient rule as i equated the fraction they give as the answer and my answer and it doesn't cancel...

This is how i set up the quotient rule:

1=2timesdy/dx(x^2-y)-2y(2x-dy/dx)(all over)(x^2-y)^2
8. (Original post by 111davey1)
Oh ok, it should simplify to the same expression though shouldn't it? i cant get that to happen.
One thing that you sometimes need to do is use the original expression (that you were differentiating).

In this case, assuming you diffed the fraction correctly, you've now got some expression with as the denominator.

You can use the original equation to replace with something a bit less unpleasant to work with. (No idea how much it will help, but I think it will).
9. (Original post by 111davey1)
i dont have a way of uploading it . But i think the problem lies with my quotient rule as i equated the fraction they give as the answer and my answer and it doesn't cancel...

This is how i set up the quotient rule:

1=2timesdy/dx(x^2-y)-2y(2x-dy/dx)(all over)(x^2-y)^2
You'd need to a lot of substituting to get to the fraction they've given but if you try numerical values, you'll see that you get the same gradient value using their fraction and yours, if you've done it correctly
10. (Original post by h3rmit)
You'd need to a lot of substituting to get to the fraction they've given but if you try numerical values, you'll see that you get the same gradient value using their fraction and yours, if you've done it correctly
11. (Original post by 111davey1)
Just multiply both sides by the denominator

expand and rearrange, you get a nice expression like that, then just implicitly differentiate like all of the other ones on the sheet before that.

no need to dive into a quotient here.
12. I was doing that exercise on friday and I just multiplied the whole equation by x^2 - y to get an equation of x^3 - xy - 2y = 0. It's much simpler that way.

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Updated: January 29, 2017
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