Visualize Itô differentiation rule Please help me to find an idea to visualize $$\displaystyle d{ f(t,x)}
=
\frac{\partial f(t,x)}{\partial t}dt
+ \frac{\partial f(t,x)}{\partial x}dx
+ \frac12 \frac{\partial^2f(t,x)}{\partial x^2}
dt$$
To clarify the question first make an example,  when we want to visualize $d(xy)$ (where x,y are function of $t$) we can use a rectangle  with $x+\Delta x ,y+\Delta y$ as long and short side,like the picture below

$$\Delta(xy)=(x+\Delta x)(y+\Delta y)-xy\\=x\Delta y+y\Delta x+\Delta x\Delta y$$so ignore term of $\Delta x\Delta y = somethimg(x)\Delta t \times somethimg(y)\Delta t=something (\Delta t)^2 \to 0$ 
   and say $d(xy)=xdy +ydx$
But in the case of Itô, $(dw_t)^2$ tends to $dt$. 
Now can you help me to find an idea to show (or visualize like rectangle example or something else )
$$\displaystyle d{ f(t,x)}
=
\frac{\partial f(t,x)}{\partial t}dt
+ \frac{\partial f(t,x)}{\partial x}dw_t
+ \frac12 \frac{\partial^2f(t,x)}{\partial x^2}
(dw_t)^2$$
 A: Notice that a full expansion to second order of the taylor derivative  would be
$df(t,x)=\frac{\partial f(t,x)}{\partial t}dt+ \frac{\partial f(t,x)}{\partial x}dw_t+ \frac12 \frac{\partial^2f(t,x)}{\partial x^2}(dw_t)^2+\color{red}{\frac12 \frac{\partial^2f(t,x)}{\partial t^2}(dt)^2+2\frac12\frac{\partial^2f(t,x)}{\partial t\partial x}(dt)(dw_t)}$
Lets rewrite this for $(x(t),y(t))$ instead of $(t,x(t))$
$df(x,y)=\frac{\partial f(x,y)}{\partial x}dw_t+\frac{\partial f(x,y)}{\partial y}dv_t+\frac12 \frac{\partial^2f(x,y)}{\partial x^2}(dw_t)^2+\frac12 \frac{\partial^2f(x,y)}{\partial y^2}(dv_t)^2+2\frac12\frac{\partial^2f(x,y)}{\partial x\partial y}(dw_t)(dv_t)$
Next we reorder the terms on those with and without a factor of $\frac12$
$df(x,y)=\underbrace{\underbrace{(\underbrace{\frac{\partial f(x,y)}{\partial x}dw_t+\frac{\partial f(x,y)}{\partial y}dv_t}_{\text{first order derivative of (x,y)}}+\frac{\partial^2f(x,y)}{\partial x\partial y}(dw_t)(dv_t))}_{\text{first order derivative of x$\cdot$y}}+\frac12( \frac{\partial^2f(x,y)}{\partial x^2}(dw_t)^2+ \frac{\partial^2f(x,y)}{\partial y^2}(dv_t)^2)}_{\text{second order derivative of (x,y)}}$
Then we can say that the first order derivative of $x\cdot y$ (or the "parallelogram" derivative) is just
$df(x,y)=\frac{\partial f(x,y)}{\partial (xy)}d(w_tv_t)=\frac{\partial f(x,y)}{\partial x}dw_t+\frac{\partial f(x,y)}{\partial y}dv_t+\frac{\partial^2f(x,y)}{\partial x\partial y}(dw_t)(dv_t)$
Finally if you let f(x,y)=xy then
$d(xy)=1\cdot d(w_tv_t)=ydw_t+xdv_t+(dw_t)(dv_t)$
So now we can visualize $df(x,y)$ while being consistent with what we already knew how to visualize $d(xy)$
