I'm always having difficulties with rather complicated derivatives, simply because I always make small, stupid mistakes in the process.
Would someone be so kind to help me with the second derivative of $y$ in terms of $y$ and $x$ ?
I did the first: $\frac{d}{dx}1 = \frac{d}{dx}x^3-\frac{d}{dx}3xy+\frac{d}{dx}y^3 = 3x^2-3y-3x\frac{dy}{dx}+\frac{dy}{dx}(3y^2)$ $\Rightarrow \frac{dy}{dx} = \frac{x^2-y}{x-y^2}$
I got totally lost with the second one though. $\frac{d^2y}{dx^2} = \frac{d}{dx}\frac{x^2}{x-y^2}-\frac{d}{dx}\frac{y}{x-y^2}=\frac{(2x)(x-y^2)-x^2(1-\frac{dy}{dx}(2y))}{(x-y^2)^2}-\frac{\frac{dy}{dx}(x-y^2)-y(1-\frac{dy}{dx}(2y))}{(x-y^2)^2}$
And this is basically where I'm loosing it.. I did substitute $\frac{dy}{dx}$ with my first result, but it ended in huge chaos, pretty far from the result wolfram-alpha suggests.
Any help would be appreciated! Also maybe any hints & tricks to avoid such monster-equations (if possible). Because that way I always unnecessarily screw up any test / exam..
Thank you.