Given the following transformations, find the differentials of the given functions
\begin{equation} \left.\begin{array}{cl}{r^{\prime}=\frac{r}{\sqrt{\left(1-r^{2} / R^{2}\right)}} e^{-ct / R},} & {t^{\prime}=t+\frac{R}{2 c} \ln \left(1-\frac{r^{2}}{R^{2}}\right)} \\ {\theta^{\prime}=\theta,} & {\phi^{\prime}=\phi}\end{array}\right\} \end{equation}
Solving for r and t we find that \begin{equation} r = r^{\prime} \left( \sqrt{1-\frac{r^{2}}{R^{2}}}\right) e^{ct/R} \end{equation} and \begin{equation} t=t^{\prime} - \frac{R}{2 c} \ln{ \left(1-\frac{r^{2}}{R^{2}}\right)} \end{equation} then the differential for $dr$ is as follows, \begin{gather*} dr = \frac{\partial r}{\partial r^{\prime}} dr^{\prime} + \frac{\partial r}{\partial t^{\prime}} dt^{\prime} \\ \implies \frac{\partial }{\partial r^{\prime}} \bigg(r^{\prime} \left( \sqrt{1-\frac{r^{2}}{R^{2}}}\right) e^{ct/R} \bigg)dr^{\prime} + 0dt^{\prime} \\ \implies \left( \sqrt{1-\frac{r^{2}}{R^{2}}}\right) e^{ct/R} dr^{\prime} \end{gather*}
which I believe is correct, nonetheless, when I attempt to derive the differential for $dt^{\prime}$ I unfortunately don't arrive at the correct differential, i.e. \begin{gather*} dt = \frac{\partial t}{\partial r^{\prime}} dr^{\prime} + \frac{\partial t}{\partial t^{\prime}} dt^{\prime} \\ \implies = \frac{\partial }{\partial r^{\prime}} \bigg( t^{\prime} - \frac{R}{2 c} \ln{ \left(1-\frac{r^{2}}{R^{2}}\right)} \bigg)dr^{\prime} + \frac{\partial }{\partial t^{\prime}} \bigg( t^{\prime} - \frac{R}{2 c} \ln{ \left(1-\frac{r^{2}}{R^{2}}\right)} \bigg)dt^{\prime} \\ \implies = 0 dr^{\prime} + (1) dt^{\prime} \end{gather*} Yet I believe that the correct differential should be something along the lines of \begin{equation} dt = \sqrt{\frac{1}{\left(1 - \frac{r^{2}}{R^{2}} \right)}} dt^{\prime} \end{equation}
Is there something that I am doing wrong? Should I define the $r^{2}$ inside the natural logarithm in terms of $r^{\prime}$? Any help would be appreciated