1. ## second derivative

bk2 p50 ex17c q20
given
$\displaystyle x^3 - 3xy +y^3 =4$
find
$\displaystyle \frac{d^2 y }{dx^2}$

my working:
i use y' for dy/dx
$\displaystyle y' = \frac {x^2-y}{x-y^2}$
$\displaystyle y''= \frac {(x-y^2)(2x-y')-(x^2-y)(1-2y y')}{(x-y^2)^2}$
$\displaystyle y''= \frac {(-x+2x^2-y^2)y'+x^2-2xy+y}{(x-y^2)^2}$
stuck here. thanks

p.s. why does that 'dash' -- y' look so bad in latex?

2. Originally Posted by afeasfaerw23231233
bk2 p50 ex17c q20
given
$\displaystyle x^3 - 3xy +y^3 =4$
find
$\displaystyle \frac{d^2 y }{dx^2}$

my working:
i use y' for dy/dx
$\displaystyle y' = \frac {x^2-y}{x-y^2}$
$\displaystyle y''= \frac {(x-y^2)(2x-y')-(x^2-y)(1-2y y')}{(x-y^2)^2}$
$\displaystyle y''= \frac {(-x+2x^2-y^2)y'+x^2-2xy+y}{(x-y^2)^2}$
stuck here. thanks

p.s. why does that 'dash' -- y' look so bad in latex?
I haven't checked your work, but the next step is to insert your expression for y' in terms of x and y. It will be messy.

-Dan

3. Hello,

Originally Posted by afeasfaerw23231233
bk2 p50 ex17c q20
given
$\displaystyle x^3 - 3xy +y^3 =4$
find
$\displaystyle \frac{d^2 y }{dx^2}$

my working:
i use y' for dy/dx
$\displaystyle y' = \frac {x^2-y}{x-y^2}$
$\displaystyle y''= \frac {(x-y^2)(2x-y')-(x^2-y)(1-2y y')}{(x-y^2)^2}$
$\displaystyle y''= \frac {(-x+2x^2-y^2)y'+x^2-2xy+y}{(x-y^2)^2}$
stuck here. thanks

p.s. why does that 'dash' -- y' look so bad in latex?
It's ok for the first derivative.

For the second derivative, I'd keep the formula you got by implicit differentiation :

$\displaystyle x^2-y-xy'+y'y^2=0$

Derivating :

$\displaystyle 2x-y'-y'-xy''+2y'^2y+y''y^2=0$

$\displaystyle 2x-2y'-xy''+2y'^2y+y''y^2=0$

And then replacing...
It's ugly, but it can be more friendly than using the quotient rule