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:
$\frac{d^2y}{dx^2} = \frac{-1}{2y+x}(\frac{1}{(1+x)^2} + 2\frac{dy}{dx} + 2(\frac{dy}{dx})^2)$
 A: 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
$y'' = -{1 \over 2y+x} ({1 \over (1+x)^2} +2(y')^2 +2y')$.
A: Another method: consider your equation
$$y^2 + xy - \ln (x+1)=0$$
as a quadratic in $x$ giving:
$$y=\frac12 \left(-x \pm \sqrt{x^2+4 \ln(x+1)}\right)\tag{1}$$
Then differentiate twice.
In an alternate way, write (1) under the equivalent form:
$$(2y+1)^2=(x^2+4 \ln(x+1))^2$$
then differentiate twice,
giving simpler calculations.
A: To follow the path you took, applying the Product Rule to
$$ \frac{dy}{dx} \ \ = \ \ \frac{1}{2y+x} · \left(\frac{1}{x+1} -y \right) \ \ = \ \ (2y + x)^{-1} \ · \ \left(\frac{1}{x+1} -y \right) $$
[which means needing to bring in the Chain Rule as well] produces
$$ \frac{d^2y}{dx^2} \ \ = \ \  \left[ \ -(2y + x)^{-2} \ · \ \left(2·\frac{dy}{dx} \ + \ 1 \right)   \    \right] \ · \ \left( \frac{1}{x+1} \ - \ y \right)   $$ $$ + \ \  (2y + x)^{-1} \ · \ \left[ \ \frac{-1}{(x+1)^2} \ - \ \frac{dy}{dx} \ \right] \ \ ,  $$
so what you have to this point appears to be correct, but could be simplified (somewhat).  The person writing the answer key "extracted" a common factor among the terms to obtain
$$ \frac{d^2y}{dx^2} \ \ = \ \ -(2y + x)^{-1} \ · \ \left[ \ \frac{1}{(x+1)^2} \ + \ \frac{dy}{dx} \ + \ (2y + x)^{-1}  ·  \left(2·\frac{dy}{dx} \ + \ 1 \right)  ·  \left( \frac{1}{x+1} \ - \ y \right)  \    \right] \ \ ,   $$
which can be re-arranged as
$$ \frac{d^2y}{dx^2} \ \ = \ \ -(2y + x)^{-1} \ · \ \left[ \ \frac{1}{(x+1)^2} \ + \ \frac{dy}{dx} \ + \  \left(2·\frac{dy}{dx} \ + \ 1 \right)  · \underbrace{(2y + x)^{-1}  ·   \left( \frac{1}{x+1} \ - \ y \right)}_{dy/dx}  \    \right] \ \ ,   $$
in which we notice that two of the factors in the last term are the expression you found for the first derivative, permitting us to write
$$ \frac{d^2y}{dx^2} \ \ = \ \ -(2y + x)^{-1} \ · \ \left[ \ \frac{1}{(x+1)^2} \ + \ \frac{dy}{dx} \ + \  \left(2·\frac{dy}{dx} \ + \ 1 \right)  · \frac{dy}{dx}  \    \right]    $$
$$ = \ \ -(2y + x)^{-1} \ · \ \left[ \ \frac{1}{(x+1)^2} \ + \ \frac{dy}{dx} \ + \  2·\left(\frac{dy}{dx}\right)^2 \ + \ \frac{dy}{dx}  \    \right]  \ \ ,  $$
leading to the given answer.
[My personal preference in finding higher derivatives by implicit differentiation is that demonstrated by copper.hat. You may need to find a simplified expression for each successive derivative, but it depends on the expression being differentiated.  It is generally a good idea to keep in mind that original expression and what you found for the derivatives, as they may be useful in making calculational simplifications.
Also, you should be more careful about your use of parentheses: you have two places in your result that could cause confusion in making the calculation. ]
