Just wondering if there is a meaning in the the following $$\frac{d^2 \sin x}{d (\cos x)^2}$$ I know that the Leibinz notation is some kind of symbolic representation of derivative. But is there any formalism behind this?
Actually let me tell the story from the beginning: I would calculate the derivative:
$$\frac{d^n}{dx^n}(1-x^2)^{n+1/2}$$ for $x=\cos kt$
Would like to find a formal way of doing this.
By definition differential for $f$ function is linear function $g(h)=A\cdot h$, for constant $A$, which represent so called principal part of the change in a function. Formally for $x_0$ point this means $$f(x_0+h)-f(x_0)=g(h)+o(h)=A\cdot h +o(h),h \to 0$$ Proved, that existence of differential is equivalence to function differentiability and $A=f^{'}(x_0)$.
For linear function $g$ is accepted notation $df(x_0) = f^{'}(x_0)dx$, where $dx$ is differential for identity function: $dx(h)=h$. So exact definition should be written: $$df(x_0)(h) = f^{'}(x_0)dx(h)=f^{'}(x_0) h$$ In brought example $d (\cos x)^2 = -2\cos x \sin x dx$
High order differential, omitting some details, is defined for simple variable as $$d^n y=y^{(n)}dx^n$$ So for brought example $d^2 \sin x=-\sin x dx^2$
Note: When I finished first part of question, I found, that you added second one starting with word "Actually". For composition situation is little more difficult and, for example, for composition $z=z[y(x)]$ second differential is $$d^2z=z^{''}_{yy}dy^2+z^{'}_{y}d^2y$$ Let me write answer to your example little later.
Addition.
As promised: $$ \begin{array}{}\frac{dy}{dx} = -(n+1)x\left(1-x^2 \right)^{\frac{n-1}{2}} \\ \frac{d^2y}{dx^2} = (n^2-1)x^2\left(1-x^2 \right)^{\frac{n-3}{2}}-(n-1)\left(1-x^2 \right)^{\frac{n-1}{2}} \\ \frac{d^3y}{dx^3} = -(n+1)n^2x^3\left(1-x^2 \right)^{\frac{n-5}{2}}+4nx^3\left(1-x^2 \right)^{\frac{n-5}{2}}-\\ -3(n+1)x^3\left(1-x^2 \right)^{\frac{n-5}{2}}+3nx(n+1)\left(1-x^2 \right)^{\frac{n-3}{2}}-\\ -3x(n+1)\left(1-x^2 \right)^{\frac{n-3}{2}} \end{array}$$ Sorry, yet, do not see general form.
Meanwhile, if you put $x=\cos kt $, then, whole expression, obviously, become more easy $y(t)=\left(\sin kt\right)^{(n+1)}$, but how it will simplify work for $\frac{d^ny}{dx^n}$, I do not see yet ,
If you would like have $\frac{d^ny}{dt^n}$ then What is the $n$th derivative of $\sin(x)$? help.
As
$$\dfrac{d^2y}{dx^2}=\dfrac{d\left(\dfrac{dy}{dx}\right)}{dx}$$
$$\dfrac{d^2(\sin x)}{d^2(\cos x)}=\dfrac{d\left(\dfrac{d(\sin x)}{d(\cos x)}\right)}{d(\cos x)}$$
Finally as $\dfrac{d(\cos x)}{dx}=-\sin x, d(\cos x)=?$
Well, $\mathrm{d}(\cos x) = - \sin x\, \mathrm{d}x$ Thus \begin{align} \frac{\mathrm{d}^2(\sin x)}{\mathrm{d}(\cos x)^2} &= \frac{\mathrm{d}}{\mathrm{d}(\cos x)}\frac{\mathrm{d} \sin x}{-\sin x \,\mathrm{d}x} \\ &=\frac{\mathrm{d}}{\mathrm{d}(\cos x)}\left(\frac{-1}{\sin x}\right)\frac{\mathrm{d} \sin x}{\mathrm{d}x} \\ &=\frac{\mathrm{d}}{\mathrm{d}(\cos x)}\left(\frac{-\cos x}{\sin x}\right) \\ &= \frac{\mathrm{d}}{-\sin x \,\mathrm{d}x}\left(\frac{-\cos x}{\sin x}\right)\\ &=\operatorname{cosec} x\frac{\mathrm{d}}{\mathrm{d}x}(\cot x)\\ &= \operatorname{cosec} x \cdot (- \operatorname{cosec}^2 x) \\ &= - \operatorname{cosec}^3 x \end{align}
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\totald{}{x} = \totald{\cos\pars{x}}{x}\,\totald{}{\cos\pars{x}} = -\sin\pars{x}\,\totald{}{\cos\pars{x}} \\[5mm] \implies &\ \totald{}{\cos\pars{x}} = -\,{1 \over \sin\pars{x}}\totald{}{x} \\[5mm] \implies &\ \totald[2]{}{\pars{\cos\pars{x}}} = {1 \over \sin\pars{x}}\totald{}{x} {1 \over \sin\pars{x}}\totald{}{x} = \bbx{-\,{\cos\pars{x} \over \sin^{3}\pars{x}}\totald{}{x} + {1 \over \sin^{2}\pars{x}}\totald[2]{}{x}} \end{align}
