Given, $$g(a)=\lim _\limits{h \rightarrow 0^{+}} \int_h^{1-h} t^{-a}(1-t)^{a-1} d t$$ is differentiable for all $$a \in(0,1)$$.

$$\therefore \quad g^{\prime}\left(\frac{1}{2}\right)=\lim _\limits{\beta \rightarrow 0} \frac{g\left(\frac{1}{2}+\beta\right)-g\left(\frac{1}{2}\right)}{\beta}$$

$$\begin{aligned} & \therefore g^{\prime}\left(\frac{1}{2}\right)=\lim _{\beta \rightarrow 0} \frac{g\left(\frac{1}{2}+\beta\right)-g\left(\frac{1}{2}\right)}{\beta} \\ & \Rightarrow g^{\prime}\left(\frac{1}{2}\right)=\lim _{\beta \rightarrow 0} \frac{1}{\beta} \\ & {\left[\lim _{h \rightarrow 0^{+}} \int_h^{1-h}\left(t^{-\left(\frac{1}{2}+\beta\right)}(1-t)^{\frac{1}{2}+\beta-1}-t^{\frac{-1}{2}}(1-t)^{\frac{-1}{2}}\right) d t\right]} \end{aligned}$$

$$\begin{aligned} \Rightarrow g^{\prime} & \left(\frac{1}{2}\right)=\lim _{h \rightarrow 0^{+}} \\ & {\left[\int_h^{1-h} t^{\frac{-1}{2}}(1-t)^{\frac{-1}{2}} \cdot\left(\lim _{\beta \rightarrow 0} \frac{t^{-\beta} \cdot(1-t)^\beta-1}{\beta}\right)\right] } \end{aligned}$$

$$\begin{aligned} \Rightarrow g^{\prime}\left(\frac{1}{2}\right) & =\lim _{h \rightarrow 0^{+}} \\ & {\left[\int_h^{1-h} t^{\frac{-1}{2}}(1-t)^{\frac{-1}{2}} d t\left(\lim _{\beta \rightarrow 0} \frac{\left(\frac{1-t}{t}\right)^\beta-1}{\beta}\right)\right] } \end{aligned}$$

$$\Rightarrow g^{\prime}\left(\frac{1}{2}\right)=\lim _\limits{h \rightarrow 0} \int_h^{1-h} t^{\frac{-1}{2}}(1-t)^{\frac{-1}{2}} d t \log _e\left(\frac{1-t}{t}\right) \quad \text{... (i)}$$

Apply the property $$\int_a^b f(x) d x=\int_a^b f(a+b-x) d x$$

$$\begin{aligned} & \Rightarrow g^{\prime}\left(\frac{1}{2}\right)=\lim _{h \rightarrow 0} \int_h^{1-h}(1-t)^{\frac{-1}{2}} t^{\frac{-1}{2}} d t \cdot \log _e\left(\frac{t}{1-t}\right) \\ & \Rightarrow g^{\prime}\left(\frac{1}{2}\right)=\lim _{h \rightarrow 0} \int_h^{1-h} t^{\frac{-1}{2}} \cdot(1-t)^{\frac{-1}{2}} d t \cdot \log _e\left(\frac{1-t}{t}\right)^{-1} \\ & \Rightarrow g^{\prime}\left(\frac{1}{2}\right)=-\lim _{h \rightarrow 0} \int_h^{1-h} t^{\frac{-1}{2}}(1-t)^{\frac{-1}{2}} d t \cdot \log _e\left(\frac{1-t}{t}\right) \\ & \Rightarrow g^{\prime}\left(\frac{1}{2}\right)=-g^{\prime}\left(\frac{1}{2}\right)\\ \begin{aligned} & \Rightarrow 2 g^{\prime}\left(\frac{1}{2}\right)=0 \\ & \Rightarrow g^{\prime}\left(\frac{1}{2}\right)=0 \end{aligned} \end{aligned}$$

Hint:

(i) If $$f(x)$$ is differentiable at $$x=a$$, then

$$f^{\prime}(a)=\lim _{\beta \rightarrow 0} \frac{f(a+\beta)-f(a)}{\beta}$$

(ii) $$\lim _\limits{x \rightarrow 0} \frac{a^x-1}{x}=\ln a$$

(iii) Recall the property

$$\int_a^b f(x) d x=\int_a^b f(a+b-x) d x$$