Answer
$$ y^{\prime}= \frac{5^{\theta} \ln 5\left(2-\log _{5} \theta\right)(\theta \ln 5+1)+5^{\theta}}{\ln 5\left(2-\log _{5} \theta\right)^{2}}$$
Work Step by Step
Given $$ y = \frac{\theta .5^{\theta}}{2-\log _{5} \theta}$$
Since $$\log_{a}z=\frac{\ln z}{\ln a}$$ So, we have
\begin{aligned}
y& = \frac{\theta .5^{\theta}}{2-\log _{5} \theta}\\
&=\frac{\theta .5^{\theta}}{2-\frac{\ln\theta}{\ln 5}} \\
&\Rightarrow y^{\prime}=\frac{\left(2-\frac{\ln \theta }{\ln 5}\right)\left(\theta \cdot 5^{\theta} \ln 5+5^{\theta}(1)\right)-\left(\theta \cdot 5^{\theta}\right)\left(-\frac{1}{\theta \ln 5}\right)}{\left(2-\frac{\ln \theta}{\ln 5}\right)^{2}}\\
&\ \ \ \ \ \ \ \ \ \ \ =\frac{(\ln 5) \ \left(2-\frac{\ln \theta }{\ln 5}\right)\left(\theta \cdot 5^{\theta} \ln 5+5^{\theta} \right)+(\ln 5) \ \left(\theta \cdot 5^{\theta}\right)\left(\frac{1}{\theta \ln 5}\right)}{(\ln 5) \left(2-\frac{\ln \theta}{\ln 5}\right)^{2}}\\
&\ \ \ \ \ \ \ \ \ \ \ =\frac{(\ln 5) \ \left(2- \log_5 \theta \right)\left(\theta \cdot 5^{\theta} \ln 5+5^{\theta} \right)+5^{\theta} }{(\ln 5) \left(2- \log_5 \theta \right)^{2}}\\
&\ \ \ \ \ \ \ \ \ \ \ =\frac{5^{\theta} \ln 5\left(2-\log _{5} \theta\right)(\theta \ln 5+1)+5^{\theta}}{\ln 5\left(2-\log _{5} \theta\right)^{2}}\\
\end{aligned}
