Parallel to line on $f(x)=1+\sin(x)/x$ I want to draw a curve on the top of the function $f(x)=1+\dfrac{\sin(x)}{x}$, but the curve should be equidistant (perpendicular distance from any point of the function $f(x)=1+\dfrac{\sin(x)}{x}$) from every point of the curve. Could anyone help in this regard? 

I've arbitrarily decided to be indulgent today, so I'll write out in full how one might generate the parallel curve of a function. Adapting the formulae from the MathWorld link to the case of a function curve, we have the following parametric equations for a parallel curve to $y=f(x)$ with offset $h$:
$$\begin{align*}
x&=t\pm\frac{h f^\prime(t)}{\sqrt{1+f^\prime(t)^2}}\\
y&=f(t)\mp\frac{h}{\sqrt{1+f^\prime(t)^2}}
\end{align*}$$
For the OP's function, we have, after some simplification,
$$\begin{align*}
x&=t\pm\frac{h (t\cos\,t-\sin\,t)}{\sqrt{t^4+(t\cos\,t-\sin\,t)^2}}\\
y&=1+\frac{\sin\,t}{t}\mp\frac{ht^2}{\sqrt{t^4+(t\cos\,t-\sin\,t)^2}}
\end{align*}$$
Here's a plot of a bunch of parallels, with the starting curve marked in red:

As I've previously noted, for certain values of $h$, the curves are no longer functions, and start displaying cusps. On the other hand, even for the parallels that are still functions, it is rather difficult to obtain an explicit $y=f(x)$ expression, so you may have to be content with the parametric representation.
