How to solve $\frac{\partial{\rm B}}{\partial b}\left(0^+,1\right)=-\frac{\pi^2}{6}$
Knowing that $$\text{B}\,(a,b)=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$$ and $$\psi(x) =\frac{d}{dx} \ln{\Gamma(x)}= \frac{\Gamma'(x)}{\Gamma(x)}\quad\Longrightarrow\quad\Gamma'(x)=\psi(x)\Gamma(x)$$ where $\Gamma(x)$ is gamma function and $\psi(x)$ is digamma function, then $$\begin{align}\require\cancel\frac{\partial\text{B}}{\partial b}&=\frac{[\Gamma(a)\Gamma(b)]'\cdot\Gamma(a+b)-\Gamma'(a+b)\cdot\Gamma(a)\Gamma(b)}{\Gamma^2(a+b)}\\&=\frac{[\Gamma'(a)\Gamma(b)+\Gamma(a)\Gamma'(b)]\Gamma(a+b)-\Gamma'(a+b)\cdot\Gamma(a)\Gamma(b)}{\Gamma(a+b)\Gamma(a+b)}\\&=\frac{[0+\Gamma(a)\psi(b)\Gamma(b)]\cancel{\Gamma(a+b)}-\psi(a+b)\cancel{\Gamma(a+b)}\cdot\Gamma(a)\Gamma(b)}{\Gamma(a+b)\cancel{\Gamma(a+b)}}\\&=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}\big[\psi(b)-\psi(a+b)\big]\\&=\text{B}\,(a,b)\big[\psi(b)-\psi(a+b)\big]\end{align}$$ Hence $$\begin{align}\frac{\partial\text{B}}{\partial b}\left(0^+,1\right)&=\lim_{a\to0^+}\text{B}\,\left(a,1\right)\big[\psi(1)-\psi(a+1)\big]\\&=\lim_{a\to0^+}\frac{1}{a}\left[-\gamma+\gamma -\sum_{k=1}^\infty (-1)^{k+1}\zeta (k+1)\;a^k\right]\tag{1}\\&=-\lim_{a\to0^+}\sum_{k=1}^\infty (-1)^{k+1}k\,\,\zeta (k+1)\;a^{k-1}\tag{2}\\&=-\lim_{a\to0^+}\left(\zeta(2)-2\zeta(3)a+3\zeta(4)a^2-4\zeta(5)a^3+\cdots\right)\tag{3}\\&=-\frac{\pi^2}{6}\tag{4}\end{align}$$
Explanation :
$(1)$ Use series representations of $\displaystyle\text{B}\,\left(x,1\right)=\frac{1}{x}$ and $\displaystyle\psi(x+1)=-\gamma +\sum_{k=1}^\infty (-1)^{k+1}\zeta (k+1)\;x^k$ for $|x|<1$
$(2)$ Use L'Hôpital's rule because of indeterminate form $\dfrac{0}{0}$
$(3)$ Expanding the series form
$(4)$ Use $\zeta(2)=\dfrac{\pi^2}{6}$
$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{#c00000}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ With $\ds{a\ >\ 0}$ and $\ds{b\ >\ 0}$:
$$ {\rm B}\pars{a,b}\equiv\int_{0}^{1}t^{a - 1}\pars{1 - t}^{b - 1}\,\dd t $$
$$ \partiald{{\rm B}\pars{a,b}}{b} =\int_{0}^{1}t^{a - 1}\pars{1 - t}^{b - 1}\ln\pars{1 - t}\,\dd t $$
\begin{align} \left.\partiald{{\rm B}\pars{a,b}}{b}\right\vert_{\, b\ =\ 1} &=\int_{0}^{1}t^{a - 1}\ln\pars{1 - t}\,\dd t =\int_{0}^{1}t^{a - 1}\bracks{-\sum_{n\ =\ 1}^{\infty}{t^{n} \over n}}\,\dd t \\[5mm]&=-\sum_{n\ =\ 1}^{\infty}{1 \over n}\int_{0}^{1}t^{n + a - 1}\,\dd t =-\sum_{n\ =\ 1}^{\infty}{1 \over n\pars{n + a}} \end{align}
$$\color{#66f}{\large% \lim_{a\ \to\ 0^{+}}\left.\partiald{{\rm B}\pars{a,b}}{b}\right\vert_{\, b\ =\ 1}} =-\sum_{n\ =\ 1}^{\infty}{1 \over n^{2}} =\color{#66f}{\large -\,{\pi^{2} \over 6}} $$