How To Find D2y Dx2

y + 2x (dy/dx) = x - 2y

diff wrt x

How do I practise this - to go d2y/dx2 do I dispense equation and do product dominion.

This is to calculate the nature of the stationary points.

Stationary points at (4,2) and (-4,-2).

I know what to do in the later stages, just how do I differentiate a differential?

Subsequently doing diff wrt ten I get: d2y/dx2 + ii dy/dx = 1 - (two * dy/dx).

Can someone please explain where I have gone incorrect, I think it is to exercise with the LHS.

Thanks

(Original post past Moordland)
y + 2x (dy/dx) = x - 2y

unequal wrt ten

How practise I practice this - to go d2y/dx2 practise I manipulate equation and do product rule.

This is to calculate the nature of the stationary points.

Stationary points at (iv,two) and (-4,-ii).

I know what to practise in the subsequently stages, but how do I differentiate a differential?

After doing diff wrt x I get: d2y/dx2 + 2 dy/dx = 1 - (2 * dy/dx).

Can someone delight explain where I have gone wrong, I think information technology is to do with the LHS.

Thanks

What happened to the kickoff y when you differentiated? You seemed to have ignored it and started past differentiating $2x\frac{dy}{dx}$ .

(Original post by notnek)
What happened to the kickoff y when you lot differentiated? You seemed to have ignored it and started past differentiating $2x\frac{dy}{dx}$ .

Sad I probably should have added brackets:

I meant to write (dy/dx) [ y+2x] = x-2y

Find d2y/dx2 should I use product rule?

(Original post by Moordland)
Sorry I probably should have added brackets:

I meant to write (dy/dx) [ y+2x] = 10-2y

Detect d2y/dx2 should I use product rule?

Y'all should divide by (y+2x) so you lot have (dy/dx) = (ten-2y)/(y+2x) and differentiate with caliber rule

(Original post by Moordland)
Sorry I probably should have added brackets:

I meant to write (dy/dx) [ y+2x] = ten-2y

Find d2y/dx2 should I use product rule?

yup product rule. Or if you don't desire to have dy/dx in the answer, so y'all could divide by [y+2x] and the states the quotient rule.

(Original post past KitikatStrider)
You should divide by (y+2x) and then you have (dy/dx) = (x-2y)/(y+2x) and differentiate with quotient rule

(Original post by gagafacea1)
yup production rule. Or if you don't want to have dy/dx in the respond, then you lot could divide by [y+2x] and us the quotient rule.

Cheers guys, I will give information technology a go!

(Original post by KitikatStrider)
Yous should divide by (y+2x) so yous take (dy/dx) = (ten-2y)/(y+2x) and differentiate with quotient rule

(Original mail service by gagafacea1)
yup product rule. Or if y'all don't want to take dy/dx in the answer, and so you lot could dissever by [y+2x] and us the quotient rule.

Non to exist a pain in the ass merely can you guys tell me where I went wrong.

http://imgur.com/5gZffmu

I go d2y/dx2 = 1 and -1 and it should exist 1/x and -one/10 respectively :cry:

(Original post by Moordland)
Not to be a hurting in the ass just can you guys tell me where I went wrong.

http://imgur.com/5gZffmu

I get d2y/dx2 = 1 and -ane and it should be 1/10 and -one/10 respectively :cry:

To start ... When you differentiated v you lot got +iii instead of +ii

I could not be bothered to bank check your multiplication of brackets every bit information technology is totally unnecessary ... You but demand to put your values into the differentiated scrap ... Since dy/dx is cipher it should be easy

Then you have some odd bit where yous have prepare the numerator to equal cypher ... That was nonsense ..

(Original mail past TenOfThem)
To offset ... When yous differentiated v you got +3 instead of +2

I could not be bothered to check your multiplication of brackets as it is totally unnecessary ... You just need to put your values into the differentiated bit ... Since dy/dx is zero it should exist like shooting fish in a barrel

Then you lot have some odd bit where you lot have ready the numerator to equal goose egg ... That was nonsense ..

Wow can't believe that.

Thank you

I eventually got the correct answers