$\frac{d^2V}{dq^2}$ of $3V^2+2q=2Vq$ As the question states I am trying to find $\frac{d^2V}{dq^2}$ of $3V^2+2q=2Vq$. I managed the first derivative and got: $$\frac{dV}{dq}=\frac{V-1}{3V-q}$$
However, when I did the second derivative I got: $$\frac{d^2V}{dq^2}=\frac{(V-1)(3V-2q+3)}{(3V-q)^3}$$
Which may or may not be right. What I am wondering about is: the book gives the answer $$\frac{d^2V}{dq^2}=\frac{(2q-3V-3)}{(1-V)^2}$$
How can I simplify my answer to get what they got, or in the case that my answer is completely wrong, could someone explain how to get to their answer.
 A: If you differentiate
$$ 3VV'+1=V+qV', $$
(from which your $V'$ follows) you get
$$ 3(V')^2+3VV''=2V'+qV'' $$
$$
\eqalign{
V''
&=\frac{V'(2-3V')}{3V-q}\\
&=\frac{(V-1)(2-3V')}{(3V-q)^2}\\
&=\frac{(V-1)(3V+3-2q)}{(3V-q)^3}.\\
}
$$
You did it right. The question is whether $V-1=3V-q$ (or their cubes are equal, which is the same thing since $y=x^3$ is invertible).
Now you can use the quadratic equation to represent $V$ in terms of $q$, but if you work it out, it's completely different. I think the book solution is either wrong, or, if you notice, their formula is really $$-\frac{V''}{(V')^3}.$$ Have they told you that $V'=-1$ by any chance?
A: I think your book is wrong. I got the same first and second derivatives as you did.
Now solving for $V$ from the original equation, we get
$$V_{1,2} = \frac{q \pm \sqrt{q^2 - 6q}}{3}.$$
Let's try $q = 8$. Then $V_1 = 4$ and $V_2 = 4/3$. Then according to the second derivative you, I, and bgins computed
$$V_1''(8) = \frac{(4-1)(3\cdot4+3-2\cdot8)}{(3\cdot4-8)^3} = \frac3{64}$$
$$V_2''(8) = \frac{(4/3-1)(3\cdot4/3+3-2\cdot8)}{(3\cdot4/3-8)^3} = \frac3{64}$$
According to your book's second derivative,
$$V_1''(8) = \frac{2\cdot8-3\cdot4-3}{(1-4)^2} = \frac19$$
$$V_2''(8) = \frac{2\cdot8-3\cdot4/3-3}{(1-4/3)^2} = 81 $$
This does not match, so there is no way to reconcile your book's answer with yours, no matter what additional constraint we add on $V$.
A: I get the same for $V''$ as you do. To simplify, use $2q(V-1) = 3V^2$, i.e.
$$\frac{(V-1)(3V-2q+3)}{(3V-q)^3} = \frac{3(V-1)(V+1) - 3V^2}{(3V-q)^3} = \frac{-3}{(3V-q)^3}$$
which I think looks nicer than the book's answer.
(See the other answer, something looks wrong or incomplete with the given answer.)
