How are values of the Dirichlet Beta function derivative derived?

$\beta'(-1)$ can actually be found directly from the functional equation. We have $$ \beta'(-1) = - \frac{\mathrm{d}}{\mathrm{d}s} \left[\left(\frac{2}{\pi}\right)^s \sin \left(\frac{\pi}{2} s\right) \Gamma(s) \beta(s)\right] ~ \Bigg\rvert_{s=2} \, .$$ Now note that the only non-vanishing term on the right-hand side is the one containing the derivative of the sine. Therefore $$\beta'(-1) = \left(\frac{2}{\pi}\right)^2 \frac{\pi}{2} [-\cos(\pi)] \Gamma(2) \beta(2) = \frac{2 \mathrm{K}}{\pi} \, .$$ However, this also means that $\beta'(2)$ cannot be calculated from $\beta'(-1)$.

---

$\beta'(1)$ is not as easy to find. We can start from the integral \begin{align} \frac{1}{2} \int \limits_0^\infty \frac{-\ln(x)}{\cosh(x)} \, \mathrm{d} x &= - \frac{1}{2} \frac{\mathrm{d}}{\mathrm{d}s} \int \limits_0^\infty \frac{x^{s-1}}{\cosh(x)} \, \mathrm{d} x ~ \Bigg\rvert_{s=1} = - \frac{1}{2} \frac{\mathrm{d}}{\mathrm{d}s} [2 \Gamma(s) \beta(s)] ~ \Bigg\rvert_{s=1} \\ &= \frac{\pi \gamma}{4} - \beta'(1) \, . \end{align} Since the left-hand side is a representation of Vardi's integral $$ \int \limits_0^1 \frac{-\ln(-\ln(x))}{1+x^2} \, \mathrm{d} x = \frac{\pi}{4} \ln\left(\frac{\operatorname{\Gamma}^4\left(\frac{1}{4}\right)}{4\pi^3}\right) \, ,$$ we obtain the desired result, from which the value of $\beta'(0)$ follows by the functional equation.

Various ways to evaluate the integral are discussed in this paper by Blagouchine. By the way, the author convincingly argues that the integral should actually be called Malmsten's integral, since Malmsten first computed it in 1842, which is in line with the claim from the Wikipedia page.

---

As an alternative route, here is a brief sketch of how I have calculated $\beta'(1)$ :

Plugging $z=\frac{1}{4}$ into Binet's first log-gamma formula, we obtain (after some algebra and integration by parts/substitutions) \begin{align} \ln\left(\frac{\operatorname{\Gamma}^4\left(\frac{1}{4}\right)}{4\pi^2}\right) &= 2 \ln(2) - 1 + \int \limits_0^\infty \frac{\tanh(x)}{x} \mathrm{e}^{-x} \, \mathrm{d} x \\ &\phantom{=} + \int \limits_0^\infty \left[\frac{\coth(x)}{x} -\frac{1}{\sinh^2 (x)}\right] \mathrm{e}^{-x} \, \mathrm{d} x - \int \limits_0^\infty \left[\coth(x)-\frac{1}{x}\right] \mathrm{e}^{-x} \, \mathrm{d} x \\ &\equiv 2\ln(2) - 1 + I_1 + I_2 - I_3 \, . \end{align}

We can then compute \begin{align} I_1 &= \gamma + \ln \left(\frac{\pi}{2}\right) - \frac{4}{\pi} \beta'(1) \, ,\\ I_2 &= \gamma \, ,\\ I_3 &= \gamma + \ln(2) - 1 \, \end{align} and solve for $\beta'(1)$ .

Another way to find $\beta'(1)$ is from Kummer's Fourier series for the logarithm of the gamma function: $$ \log \Gamma(x) = \left(\tfrac{1}{2}-x \right)(\gamma+\log2)+(1-x)\log \pi - \frac{1}{2} \log \sin(\pi x) + \frac{1}{\pi} \sum_{n=1}^{\infty} \frac{\log n}{n} \sin(2\pi nx) $$ valid for $0<x<1 $. Setting $x=\frac{1}{4}$ gives $$ \log \Gamma \left(\tfrac{1}{4} \right) = \frac{1}{4} (\gamma+\log 2) +\frac{3}{4}\log \pi -\frac{1}{2} \log \left(\tfrac{1}{\sqrt{2}} \right) +\frac{1}{\pi} \sum_{n=0}^{\infty} \frac{(-1)^n \log(2n+1)}{2n+1} $$ The sum appearing on the right is $-\beta'(1)$.