For which values of $d$ does $x - d\tanh(x) = 0$ have a positive solution? 
For which $d$ values does the equation $$x - d\tanh(x) = 0$$ have a positive solution?

I have tried rearranging this a number of different ways using the exponential form and and using hyperbolic trig identities but I am not entirely sure what I am looking for exactly.
Am I trying to find a positive solution for $x=d\tanh(x)$? How am I supposed to relate $d$ and $x$?
EDIT: I have a provided answer of $d = \frac{1}{\operatorname{sech}^2(x)} > 1$.
 A: The question asks you to list all the possible values for $d$ such that there exists some $x>0$ such that $x-d\tanh x=0.$ Since $\tanh$ is $0$ only at $0$, the equation arranges to $$x\coth(x)=d.$$
So basically you want to find the range of $x\coth(x)$ subject to $x>0.$ 

Edit: I'm unsure what the new question means exactly, but it seems to be: For all values of $x>0 $ such that $\displaystyle x-\frac{\tanh(x)}{\operatorname{sech}^2(x)}=0,$ return $d=\frac{1}{\operatorname{sech}^2 x}.$
A: As you've discovered, any $d>1$ suffices, as the tangent line of $\tanh(x)$ at $x=0$ is $y=x$, and the derivative of $\tanh(x)$ is strictly increasing on $(0,\infty)$. This generalizes to any function $f(x)$ with these properties, not just $\tanh(x)$.
A: Since $d=0$ is not admissible in your problem, you can remark that your equation is equivalent to $$\frac{x}{d} = \tanh x.$$
Sketch the graph of $y=\tanh x$: you are looking for the angular coefficients of those straight lines from the origin that intersect this graph at points with positive abscissae. It shouldn't be too hard to give the answer.
