Answer
$\dfrac{\ln 2}{ \sqrt{1+(0.25)^\theta}}$
Work Step by Step
Given: $y=csch^{-1} (\dfrac{1}{2})^{\theta}$
It can be re-written such as: $y=csch^{-1} (0.5)^{\theta}$ and $\dfrac{d m^x}{dx}=m^x \ln m$
Since, $\dfrac{d (csch^{-1} x)}{dx}=\dfrac{-1}{|x| \sqrt{1 + x^2}}$
Apply product rule to get the differentiation.
Thus, $\dfrac{dy}{d \theta}=\dfrac{-1}{|(0.5)^{\theta}| \sqrt{1 + ((0.5)^{\theta})^2}}(0.5)^{\theta} \ln (0.5)$
or, $\dfrac{dy}{d \theta}=\dfrac{\ln 2}{ \sqrt{1+(0.25)^\theta}}$
