Long, complicated implicit differentiation, finding $y'$ Find $y''$ where $y^2 + xy = \ln (x+1)$
Using implicit differentiation,
$\frac{dy}{dx} = \frac{1}{2y+x} (\frac{1}{x+1} -y)$
Now finding the second derivative of this function, using the product rule,
$\frac{d^2y}{dx^2} = (\frac{-2 \frac{dy}{dx} +1}{(2y+x)^2}  \times \frac{1}{x+1} - y) + (\frac{1}{2y+x} \times (\frac{-1}{(x+1)^2} - \frac{dy}{dx}))$
This looks very different from what the answer key is telling me:
$\frac{d^2y}{dx^2} = \frac{-1}{2y+x}(\frac{1}{(1+x)^2} + 2\frac{dy}{dx} + 2(\frac{dy}{dx})^2)$

Just differentiate the entire expression:
$2y y' + xy' + y = (2y+x) y' + y = {1 \over 1+x}$, then again to get
$(2y+x) y'' + (2y' + 1) y' + y' = - {1 \over (1+x)^2}$.
Rearranging gives
$y'' = -{1 \over 2y+x} ({1 \over (1+x)^2} +2(y')^2 +2y')$.
