I've been studying the shape of the cable in suspension bridges (for my school essay). I have derived a differential equation which models the shape
$y'' = m\sqrt{1 + (y')^{2}} + n$
where $m$ is the weight density of the cable over the horizontal component of the tension and $n$ is the weight density of the bridge deck over the horizontal component of the tension. I attempt to solve by moving $n$ to the LHS and squaring both sides
$(y'')^{2} - 2ny'' + n^{2} = m^{2} + m^{2}(y')^{2}$
and then taking the derivative of both sides
$2y''y''' - 2ny''' = 2m^{2}y'y''$
At this point I have no idea how to transform the equation in order to solve it besides dividing both sides by $y''$. The problem here is with the $2ny'''$ term (on the lhs). May I divide the third derivative by the second derivative in the following way?
$\frac{d}{dx}\frac{d}{dx}\frac{dy}{dx} \cdot \frac{dx}{d}\frac{dx}{dy} = \frac{d}{dx}$
Is this valid?
In Liebniz notation, the "d" is not a variable, but a function, and $d^2y$ does NOT refer to $d\cdot d\cdot y$ but rather to $d(d(y))$. Likewise, while the second derivative might be $\frac{d^2y}{dx^2}$, it is more formally $\frac{d^2y}{dx^2} - \frac{dy}{dx}\frac{d^2x}{dx^2}$, with $d^2x$ reducing to 0 if $x$ is the independent variable.
Thus, if $x$ is the independent variable, and you divide the third derivative by the second derivative, you would get:
$$\frac{d^3y}{dx^3} \cdot \frac{dx^2}{d^2y} = \frac{d^3y}{dx\,d^2y}$$
This is because the real way you should think about it is:
$$\frac{d(d(d(y)))}{(d(x))^3} \cdot \frac{(d(x))^2}{d(d(y))} = \frac{d(d(d(y)))}{d(x)\,d(d(y))}$$
If $x$ is not the independent variable, you would get a huge mess (the full expansion of higher derivatives get messier with each order).
You can't reduce the nested $d()$ function any more than you could reduce a nested $\sin()$ function.