How to show continuity and monotonicity of a solution to this parametrized equation? Let $1 \le p <2$ be a parameter. I am interested in the behaviour of solutions $s=s(p) \in (\frac{1}{4},1)$ to the equation
$$
\frac{2^{p/2} (1-\sqrt{s})^p-1}{\sqrt{s}}=-2^{p/2-1}p(1-\sqrt{s})^{p-1}. \tag{1}
$$
For $p=2$, the only solution is $s=\frac{1}{4}$, and for $p=1$ the only solution is $s=(2-\sqrt 2)^2 \simeq 0.343$ (details below).
I am rather certain that for each $1 \le p <2$, there is unique solution $s=s(p)$ in the range $(\frac{1}{4},1]$. (although I would be happy to see other arguments for that as well).

Question: Is $ p\to s(p)$ monotonically decreasing in $p$? Is it continuous in $p$? (where $1 \le p <2$).
I guess that a naive idea would be to assume $s(p)$ is differentiable, and differentiate equation $(1)$ w.r.t $p$. This seems unpleasant.

As expected Mathematica doesn't give an exact closed-form formula for the solution in terms of $p$ (although it does solve the problem for $p=1,2$).

Analysis of $p=1,2$:
For $p=1$ the equation reduces to $ \sqrt 2(1-\sqrt s)-1=-\frac{\sqrt s}{\sqrt 2}$. Setting $x=\sqrt s$, we obtain
$ 1-\sqrt 2=x(1/\sqrt 2-\sqrt 2) \Rightarrow x=2-\sqrt 2$.
For $p=2$ the equation reduces to
$2(1-\sqrt s)^2-1=-2(1-\sqrt s)\sqrt s \Rightarrow -1=-2(1-\sqrt s)\big((1-\sqrt s)+\sqrt s \big) \Rightarrow 2(1-\sqrt s)=1 \Rightarrow s=\frac{1}{4}$.

Motivation: The solutions to this equation arise naturally when trying to compute thresholds for convex envelopes of certain functions.
 A: I did exactly what you said not to do and I differentiated both sides with respect to $p$. I then solved for $s'(p)$. Of course, I didn't do this by hand. Here's the SageMath input:
p = var('p')
s = function('s')(p)

LHS = (2^(p/2)*(1-sqrt(s))^p - 1)/sqrt(s)
RHS = -2^(p/2 - 1)*p*(1-sqrt(s))^(p-1)
eqn = diff (LHS - RHS, p).full_simplify()
eqn

The output was really long, so I won't put it here. (Let me know in a comment if you'd like me to paste it). I then isolated for $s'(p)$ by first defining a function $g$ and manually replacing all the $\frac{\partial s}{\partial t}$ expressions with $g$. I typed:
g = function ('g')(p)
eqn = ... # copy and paste output from previous block and replace all instances of s' with g
soln = solve ([eqn==0], g)
soln

This gave an expression for $s'(p)$ in terms of $s$ and $p$. If you plot a streamline plot or slope field for $s'(p)$, you get this:
s = var('s') 
soln = ....    # copy and paste output from above, replacing s(p) with s
streamline_plot(soln, (p,1,2), (s,1/4,1))   


This is a stream line plot of $s(p)$: the function must follow one of these lines, depending on its initial condition. The horizontal-axis is the $p$-variable and the vertical-axis is $s$-variable. So $s'(p)$ is the slope of the lines. So $s(p)$ seems monotonically decreasing on the interval, as required. This seems to suggest that there is a unique solution for every initial condition. (Actually, this would follow from the existence and uniqueness theorem for ODEs, at least in a neighbourhood of the initial condition.)
If you want to prove this formally, you can just prove that the huge expression Sage gives you for the variable $\texttt{soln}$ ($s'(p)$) in the block above is negative. It might take an hour or so, but there's a lot of repetition so it's not too bad. All the best!
