implicit differentiation

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.
Thanks for your help:

Question:
http://pmt.physicsandmathstutor.com/download/Maths/A-level/C4/Solutionbank-Heinemann/C4%20Chapter%204.pdf

excercise b question 1f

Thanks for your help

Reply 1

DFranklin

18

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.

Reply 2

111davey1

OP

14

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.

Reply 3

h3rmit

15

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

Reply 4

111davey1

OP

14

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

Reply 5

h3rmit

15

Original post by 111davey1

I think i did.. but clearly i didnt

Can I see your working?

Reply 6

111davey1

OP

14

Original post by h3rmit

Can I see your working?

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

Reply 7

DFranklin

18

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

(x^2-y)^2

as the denominator.

You can use the original equation to replace [latex(]x^2-y)^2 with something a bit less unpleasant to work with. (No idea how much it will help, but I think it will).

Reply 8

h3rmit

15

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

Reply 9

111davey1

OP

14

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

Thanks for your help

Reply 10

PedanticStudent

13

Original post by 111davey1

Thanks for your help

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. :smile:

no need to dive into a quotient here.

(edited 7 years ago)

Reply 11

ManLike007

19

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.

implicit differentiation

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