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**Paymemoney** Hi

I need help on the following:

1) Find $\displaystyle \frac{d^{2}y}{dx^2}$ for $\displaystyle x^2 + 4xy +3y^2 = 5$ at points (-1,2)$\displaystyle

2x + 4x \frac{dy}{dx} +4y + 6y \frac{dy}{dx} = 0$

$\displaystyle \frac{dy}{dx} = \frac{-4x -2x}{6y +4x}$

Should be $\displaystyle \frac{dy}{dx} = \frac{-4y-2x}{6y+4x}$

$\displaystyle \frac{(-6y+4x)(-2 - 4\frac{d^{2}y}{dx^2}) - (-2x-4y)(6

\frac{d^{2}y}{dx^2}+4)}{(6y+4x)^2} = 0$

$\displaystyle \frac{d^{2}y}{dx^2}-24y+24y-16x+12x = 12y-16y+8x-8x$

$\displaystyle \frac{d^{2}y}{dx^2} = \frac{-4y}{-4x}$

$\displaystyle \frac{d^{2}y}{dx^2} = \frac{y}{x} * \frac{1}{(6y+4x)^2}$

when x=-1 and y=2 then:

$\displaystyle \frac{d^{2}y}{dx^2} = \frac{2}{2(128)} = \frac{1}{64}$

book's answers says: $\displaystyle \frac{5}{64}$

2) Find $\displaystyle \frac{d^{2}y}{dx^2}$ for $\displaystyle x= 1 - cos(t)$ $\displaystyle y= t - sin(t)$

$\displaystyle \frac{dx}{dt} = sin(t)$

$\displaystyle \frac{dy}{dt} = 1 - cos(t)$

$\displaystyle \frac{dy}{dx} = \frac{1 - cos(t)}{sin(t)}$

$\displaystyle \frac{d^{2}y}{dx^2} = \frac{(sin(t))^2 - cos(t) - (cos(t))^2}{(sin(t))^2}$

$\displaystyle \frac{d^{2}y}{dx^2} = \frac{1 - cos(t)}{(sin(t))^2}$

answer says it is: $\displaystyle \frac{1 - cos(t)}{(sin(t))^3}$

P.S