The sum of fractional powers $\sum\limits_{k=1}^x k^t$.
This is just a "what if ?" consideration, not an answer, and I just guess that it might be of some help to your scope. So, flanking the analysis you are conducting, you may consider this alternative development for $S_x(t)$.
- 1st development
$$
\begin{gathered}
S_x (t) = \sum\limits_{k = 1}^x {k^{\,t} } = \sum\nolimits_{\;k = 1}^{\;x + 1} {k^{\,t} } = \frac{{B_{\,t + 1} (x + 1) - B_{\,t + 1} (1)}}
{{t + 1}} = \quad \quad \left( \text{1} \right) \hfill \\
= \sum\nolimits_{\;k = 0}^{\;x} {\left( {k + 1} \right)^{\,t} } = \sum\nolimits_{\;k = 0}^{\;x} {\sum\limits_{0\, \leqslant \,j} {\left( \begin{gathered}
t \hfill \\
j \hfill \\
\end{gathered} \right)k^{\,j} } } = \sum\limits_{0\, \leqslant \,j} {\left( \begin{gathered}
t \hfill \\
j \hfill \\
\end{gathered} \right)\sum\nolimits_{\;k = 0}^{\;x} {k^{\,j} } } = \hfill \\
= \sum\limits_{0\, \leqslant \,j} {\left( \begin{gathered}
t \hfill \\
j \hfill \\
\end{gathered} \right)\left( {\frac{{B_{\,j + 1} (x) - B_{\,j + 1} (0)}}
{{j + 1}}} \right)} = \quad \quad \left( 2 \right) \hfill \\
= \sum\nolimits_{\;k = 0}^{\;x} {\sum\limits_{\begin{array}{*{20}c}
{0\, \leqslant \,j} \\
{0\, \leqslant \,l\,\left( { \leqslant \,j} \right)} \\
\end{array} } {\left( \begin{gathered}
t \hfill \\
j \hfill \\
\end{gathered} \right)\left\{ \begin{gathered}
j \\
l \\
\end{gathered} \right\}k^{\,\underline {\,l\,} } } } = \hfill \\
= \sum\limits_{\begin{array}{*{20}c}
{0\, \leqslant \,j} \\
{0\, \leqslant \,l\,\left( { \leqslant \,j} \right)} \\
\end{array} } {\left( \begin{gathered}
t \hfill \\
j \hfill \\
\end{gathered} \right)\left\{ \begin{gathered}
j \\
l \\
\end{gathered} \right\}\frac{{x^{\,\underline {\,l + 1\,} } }}
{{l + 1}}} = \sum\limits_{\begin{array}{*{20}c}
{0\, \leqslant \,j} \\
{0\, \leqslant \,l\,\left( { \leqslant \,j} \right)} \\
\end{array} } {\frac{{t^{\,\underline {\,j\,} } }}
{{j!}}\left\{ \begin{gathered}
j \\
l \\
\end{gathered} \right\}\frac{{x^{\,\underline {\,l + 1\,} } }}
{{l + 1}}} \quad \quad \left( 3 \right) \hfill \\
\end{gathered}
$$
where the symbol $\sum\nolimits_{\;k = 1}^{\;x + 1} {}$ indicates
the indefinite sum , computed between the indicated bounds,
and the curly backets the Stirling N. of 2nd kind.
To the purpose of derivating vs. $t$ and $x$ , you may replace the falling factorials $t^{\,\underline {\,j\,} } $ and$x^{\,\underline {\,l + 1\,} } $ with the corresponding Stirling devopment in $t^n$ and $x^m$ or with their expression through the Gamma function. - 2nd development
You can also write $S_x(t)$ in terms of the Hurwitz zeta function $$ \begin{gathered} S_x (t) = \sum\limits_{k = 1}^x {k^{\,t} } = \sum\nolimits_{\;k = 1}^{\;x + 1} {k^{\,t} } = \hfill \\ = \sum\nolimits_{\;k = 1}^{\;\infty } {k^{\,t} } - \sum\nolimits_{\;k = x + 1}^{\;\infty } {k^{\,t} } = \sum\nolimits_{\;k = 0}^{\;\infty } {\left( {k + 1} \right)^{\,t} } - \sum\nolimits_{\;j = 0}^{\;\infty } {\left( {j + x + 1} \right)^{\,t} } = \hfill \\ = \zeta ( - t,1) - \zeta ( - t,x + 1)\quad \quad \left( 4 \right) \hfill \\ \end{gathered} $$ - Note concerning the handling of sums and products with non-integer bounds
First let's note that $$ \begin{gathered} S_x (t) = \sum\limits_{k = 1}^x {k^{\,t} } \quad \Rightarrow \hfill \\ \Rightarrow \quad x^{\,t} = S_{x + 1} (t) - S_{x + 1} (t) = \left( {S_{x + 1} (t) + c(x + 1)} \right) - \left( {S_x (t) + c(x)} \right) \hfill \\ \end{gathered} $$ and $$ \begin{gathered} M_x (t) = \prod\limits_{k = 1}^x {k^{\,k^{\,t} } } = \prod\nolimits_{\;k = 1\;}^{\;x + 1} {k^{\,k^{\,t} } } = \prod\nolimits_{\;k = 0\;}^{\;x} {\left( {k + 1} \right)^{\,\left( {k + 1} \right)^{\,t} } } \quad \Rightarrow \hfill \\ \Rightarrow \quad \left( {x + 1} \right)^{\,\left( {x + 1} \right)^{\,t} } = \frac{{M_{x + 1} (t)}} {{M_x (t)}} = \frac{{c(x + 1)M_{x + 1} (t)}} {{c(x)M_x (t)}} \hfill \\ \end{gathered} $$ with $$ c(x)\;:\quad \text{any}\,\text{periodic}\,\text{function}\text{,}\,\text{with}\,\text{period}\,\;1 $$ Then let's take for example the starting base of your development, we get the following two different "definitions" for $B_t(x+1)$ $$ \begin{array}{*{20}c} {S_x (t) = \frac{{B_{\,t + 1} (x + 1) - B_{\,t + 1} (1)}} {{t + 1}}} \hfill & \begin{gathered} \hfill \\ = \hfill \\ \hfill \\ \end{gathered} \hfill & \begin{gathered} = \sum\limits_{k = 1}^x {k^{\,t} } = \sum\nolimits_{\;k = 1}^{\;x + 1} {k^{\,t} } = \hfill \\ = \sum\nolimits_{\;k = 0}^{\;\infty } {\left( {k + 1} \right)^{\,t} } - \sum\nolimits_{\;k = 0}^{\;\infty } {\left( {k + x + 1} \right)^{\,t} } \hfill \\ \end{gathered} \hfill \\ \hline \begin{gathered} \quad \quad \quad \quad \Downarrow \hfill \\ \frac{\partial } {{\partial \,x}}S_x (t) = \hfill \\ = \frac{1} {{t + 1}}\frac{\partial } {{\partial \,x}}B_{\,t + 1} (x + 1) = \hfill \\ = B_{\,t} (x + 1) = \hfill \\ = - t\sum\nolimits_{\;k = 0}^{\;\infty } {\left( {k + x + 1} \right)^{\,t - 1} } \hfill \\ \end{gathered} \hfill & \begin{gathered} | \hfill \\ | \hfill \\ | \hfill \\ | \hfill \\ | \hfill \\ | \hfill \\ | \hfill \\ | \hfill \\ \end{gathered} \hfill & \begin{gathered} \quad \quad \quad \quad \Downarrow \hfill \\ \frac{{B_{\,t + 1} (x + 1)}} {{t + 1}} = f(t + 1) - \sum\nolimits_{\;k = 0}^{\;\infty } {\left( {k + x + 1} \right)^{\,t} } \hfill \\ \quad \quad \quad \quad \Downarrow \hfill \\ \hfill \\ B_{\,t} (x + 1) = \hfill \\ = t\,f(t) - t\sum\nolimits_{\;k = 0}^{\;\infty } {\left( {k + x + 1} \right)^{\,t - 1} } \hfill \\ \end{gathered} \hfill \\ \end{array} $$ where - the derivate in $x$ is first taken in extending to real index the known property for integer index, and then by derivating the espression of $S(x)$ as difference of the two sums;
- $f(t)$ can be any function in $t$, and in particular it could be $B_t(1)$, which in turn can be taken as $t\;\zeta (1 - t)$, as it is in many papers concerning the extension of Bernoulli polynomials.
Thus it is evident that such mathematical entities shall be handled with great care, and specially when taking derivatives.
Applying the Euler-Maclaurin Sum Formula
The Euler-Maclaurin Sum Formula can be applied to $k^t$ to get the approximation $$ \sum_{k=1}^nk^t=\zeta(-t)+\frac1{t+1}n^{t+1}+\frac12n^t+\frac{t}{12}n^{t-1}-\frac{t^3-3t^2+2t}{720}n^{t-2}+O\left(n^{t-3}\right) $$ When $t\lt-1$, this describes how the series for $\zeta(-t)$ converges.
Possible Extension to Non-Integral Summation Limits
Consider $$ \begin{align} \lim_{\delta\to0}\frac1\delta\left(\sum_{k=1}^{n+\delta}k^t\color{#C00000}{-\sum_{k=1}^{m+\delta}k^t}\right) &=\lim_{\delta\to0}\frac1\delta\sum_{k=m+1+\delta}^{n+\delta}k^t\\ &=\lim_{\delta\to0}\frac1\delta\sum_{k=m+1}^n(k+\delta)^t\\ &=t\sum_{k=m+1}^nk^{t-1} \end{align} $$ Thus, if we give a meaning to taking a derivative with respect to the upper limit of summation, it would give $$ \frac{\mathrm{d}}{\mathrm{d}n}\sum_{k=1}^nk^t=t\sum_{k=1}^nk^{t-1}\color{#C00000}{+C} $$ where $C$ is related to the behavior near $m=0$.