Getting the benefits of exponents without exponents

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I'm trying to specify a function $f$ for a ratio of two positive variables $x_1$ and $x_2$ so that it is given by

$$R(x_1,x_2)=\frac{f(x_1)}{f(x_1)+f(x_2)}$$

And

$$R(x_2,x_1)=\frac{f(x_2)}{f(x_1)+f(x_2)}$$

One property I want $f$ to satisfy is that

$$\frac{dR(x_1,x_2)}{dx_1}/\frac{dR(x_2,x_1)}{dx_2}=\frac{x_2^c}{x_1^c}$$

Where $c$ is some positive constant. For example, I considered one that satisfies this:

$$R(x_1,x_2)=\frac{exp(x_1^{1-c})}{exp(x_1^{1-c})+exp(x_2^{1-c})}$$

(where $exp$ simply means $e$) $exp$ has the nice property of 'resisting' differentiation, so we have that the above property is satisfied. However, $exp$ makes certain kinds of solutions with optimization difficult. Is there any other kind of functional form for $f$ that would satisfy the third condition (i.e. the third equation)?


To illustrate an example that fails is simply $f(x)=x^{1-c}$, since the solution to the third equation simply is $\frac{x_2}{x_1}$ and not $\frac{x_2^c}{x_1^c}$

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Let g(x,y) = f(x)/(f(x) + f(y)), h(x,y) = f(y)/(f(x) + f(y)).

Using g_x, h_y for the partial derivates with respect to x,y.

Basically you are asking for f's that satisfy
g_x(x,y)/h_y(x,y) = x^c/y^c for some c > 0.

g_x(x,y) = f'(x)/(f(x) + f(y)) - f(x)f'(x)/(f(x) + f(y))^2
= f'(x)(f(x) + f(y) - f(x)) / (f(x) + f(y))^2
= f'(x)f(y)/(f(x) + f(y))^2
h_y(x,y) = f'(y)f(x)/(f(x) + f(y))^2.

Thus f_x(x,y)/f_y(x,y) = f'(x)f(y)/f'(y)f(x) = x^c/y^c.
x^c.f(x)/f'(x) = y^c.f(y)/f'(y). Whereupon for some constant k,
f'(x)/x^c.f(x) = k; (log f(x))' = f'(x)/f(x) = kx^c.

Solving for f, $log f(x) = kx^{c+1} / (c + 1) + d$, d in R
$f(x) = exp ((kx^{c+1} / (c + 1)) + d) = b.exp (kx^{c+1} / (c + 1)), b,c > 0$, k in R
which satisfies for all x, f(x) /= 0 /= f'(x) as used above.