<s>[INST] Simple logarithmic differentiation Given the equation $\frac{dy}{dx} = \frac{-x}{y}$
How can you solve for $\frac{d^2y}{d^2x}$ using logarithmic differentiation?
Here is my work:
$ln(\frac{dy}{dx}) = ln(\frac{-x}{y})$
$ln(\frac{dy}{dx}) = ln(-x) - ln(y)$
(Take derivative of both sides)
$(\frac{d^2y}{d^2x})/(\frac{dy}{dx}) = \frac{1}{x} - \frac{1}{y}$
(Then solve for $\frac{d^2y}{d^2x}$)
$\frac{d^2y}{d^2x} = \frac{\frac{dy}{dx}}{x} - \frac{\frac{dy}{dx}}{y}$
(Then plug in $\frac{-x}{y}$ for $\frac{dy}{dx}$)
$\frac{d^2y}{d^2x} = \frac{-x}{yx} + \frac{x}{y^2}$
And simplify to get:
$\frac{d^2y}{d^2x} = \frac{-1}{y} + \frac{x}{y^2}$
$ = \frac{(x-y)}{y^2}$
However, when I solve the derivative using the quotient rule, I get:
$\frac{x\frac{-x}{y} - y}{y^2}$
$= \frac{\frac{-x^2}{y}-y}{y^2}$
What am I doing wrong?
 [/INST] That equation can be rewritten with another notation like y'y = -x
simply claculate the dervative of that expression
$$y"y+y'²= -1 $$
where y'= dy/dx and y" = d²y/dx²
Rewriting this last equation with your notation
$$d²y/dx²=-(x²+y²)/y³$$
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