Why is $y'' = d/dy[(1/2)y'^2]$? I was wondering why the formula above is correct because if you differentiate $(1/2)y'^2$, shouldn't you get $y'.y''$ by the chain rule?
 A: It may help to think about how to interpret $\displaystyle\frac{1}{2}\frac{\mathrm{d}(y')^2}{\mathrm{d}y}$. We may define
$$ \Delta y:=y(x+\Delta x)-y(x), \qquad \Delta(y')^2:=\big[y'(x+\Delta x)\big]^2-\big[y'(x)\big]^2 $$
which depend on both $x$ and a quantity we call $\Delta x$. Note that $\Delta y$ and $\Delta x$ tend to $0$ together, so
$$ \frac{1}{2}\frac{\mathrm{d}(y')^2}{\mathrm{d}y}:=\frac{1}{2}\lim_{\Delta y\to0} \frac{\Delta (y')^2}{\Delta y} =\frac{1}{2}\lim_{\Delta x\to0} \frac{\Delta(y')^2/\Delta x}{\Delta y/\Delta x} = \frac{1}{2}\frac{\big[(y')^2\big]'}{y'} $$
Of course by chain rule, $\big[(y')^2\big]'=2y'y''$ so you get cancellation.
(This is an elaboration of user159888's answer, really.)
A: $ d/dy[(1/2)y'^2]=(d/dx [(1/2)y'^2] ) /(dy/dx)=(1/2).2y'.y''. 1/y'=y''$ where $y'=dy/dx.$
A: To me at least, it makes most sense when you use algebraically manipulable differentials.  With these, $$y'' = \frac{d^2y}{dx^2} - \frac{dy}{dx}\frac{d^2x}{dx^2}$$
To understand where this comes from, when you take the derivative of $\frac{dy}{dx}$ use the quotient rule.  See also my paper "Extending the Algebraic Manipulability of Differentials."
So, we have:
$$ \frac{d\left(\frac{1}{2}\left(\frac{dy}{dx}\right)^2\right)}{dy} \\
 \frac{2 \cdot \frac{1}{2}\left(\frac{dy}{dx}\right)\left(\frac{d^2y}{dx} - \frac{dy}{dx}\frac{d^2x}{dx}\right)}{dy} \\
 \frac{1}{dy}\left(\frac{dy}{dx}\right)\left(\frac{d^2y}{dx} - \frac{dy}{dx}\frac{d^2x}{dx}\right) \\
\frac{d^2y}{dx^2} - \frac{dy}{dx}\frac{d^2x}{dx^2}
$$
Which is the definition above of the second derivative of $y$.
A: $y$ is a function of $x$, that is $y=y(x)$; so you must derive with respect of $x$.
Hence when you think of $y'(x)^2$ you have, because of the derivative of a composition, to derive the exterior function (which is the square function) and multiply by the derivative of the interior function (which is $y'(x)$), so you have that
$$\frac{\text{d}}{{\text{d}x}}[y'(x)^2]=2y'(x)y''(x)\iff\frac{1}{2} \frac{\text{d}}{{\text{d}x}} \left[y'(x)^2\right]=y'(x)y''(x)$$
Since $\frac{1}{2}$ is constant you can bring it inside the derivative (remember the linearity of derivative), so
$$\frac{\text{d}}{\text{d}x}\left[\frac{1}{2}y'(x)^2\right]=y'(x)y''(x)$$
