What did I do wrong?
Thanks,
The first derivative is correct $\displaystyle \frac{dy}{dx}=\frac{-y^{2}}{x^{2}}$
Now, continue from there by taking the derivative of that. BUT, do not forget to sub in what dy/dx equals. That is where many go astray.
Quotient rule:
$\displaystyle \frac{x^{2}(-2y)\frac{dy}{dx}-(-y^{2})(2x)}{x^{4}}$
$\displaystyle \frac{-2y}{x^{2}}\frac{dy}{dx}+\frac{2y^{2}}{x^{3}}$
Sub in dy/dx from above:
$\displaystyle \frac{-2y}{x^{2}}(\frac{-y^{2}}{x^{2}})+\frac{2y^{2}}{x^{3}}$
$\displaystyle \frac{2y^{2}}{x^{3}}+\frac{2y^{3}}{x^{4}}$
$\displaystyle y''=\frac{2(x+y)y^{2}}{x^{4}}$
They just used a common denominator and rewrote it. That is the same thing.
Hello, r_maths!
We have: .$\displaystyle \frac{dy}{dx}\:=\:\frac{\text{-}y^2}{x^2}$$\displaystyle \frac{1}{x}+\frac{1}{y} \:=\:\frac{1}{\pi}\qquad\text{Find: }\:\frac{dy}{dx}\,\text{ and }\,\frac{d^2y}{dx^2} $
First
$\displaystyle -\frac{1}{x^2} - \frac{1}{y^2}\!\cdot\!\frac{dy}{dx} \:=\:0\quad\Rightarrow\quad \frac{dy}{dx} \:=\:-\frac{y^2}{x^2}\;\;{\color{blue}\hdots\text{ Right!}}$
Second
$\displaystyle -\frac{1}{x^2} - \frac{1}{y^2}\!\cdot\!\frac{dy}{dx} \:=\:0\;\;{\color{blue}\hdots \text{Why did you start back there?}}$
Then: . $\displaystyle \frac{d^2y}{dx^2} \;=\;\frac{x^2(\text{-}2y)\left(\frac{dy}{dx}\right) -(\text{-}y^2)(2x)}{(x^2)^2} \;=\; \frac{-2x^2y\cdot\frac{dy}{dx} + 2xy^2}{x^4}$
. . . . $\displaystyle =\;\frac{2xy\left(y - x\!\cdot\!\frac{dy}{dx}\right)}{x^4} \;=\;\frac{2y\left[y - x\left(\text{-}\frac{y^2}{x^2}\right)\right]}{x^3}$
. . . . $\displaystyle = \;\frac{2y\left(y + \frac{y^2}{x}\right)}{x^3} \;=\;\frac{2y(xy + y^2)}{x^4}\;=\;\boxed{\frac{2xy + 2y^3}{x^4}}$