Inverse Hyperbolic Functions in Terms of Logarithm

Draw the graph of ${\mathrm{cosech}}^{-1}x$. Find the value of ${\mathrm{cosech}}^{-1}\left(\frac{1}{3}\right)$. 

Draw the graph of ${\mathrm{cosech}}^{-1}x$. Find the value of ${\mathrm{cosech}}^{-1}\left(\frac{1}{2}\right)$. 

Draw the graph of ${\mathrm{cosech}}^{-1}x$. Find the value of ${\mathrm{cosech}}^{-1}\left(\frac{1}{5}\right)$. 

Draw the graph of ${\mathrm{sech}}^{-1}x$. Find the value of ${\mathrm{sech}}^{-1}\left(\frac{1}{5}\right)$. 

Draw the graph of ${\mathrm{sech}}^{-1}x$. Find the value of ${\mathrm{sech}}^{-1}\left(\frac{1}{2}\right)$. 

Draw the graph of ${\mathrm{sech}}^{-1}x$. Find the value of ${\mathrm{sech}}^{-1}\left(1\right)$. 

Draw the graph of ${\mathrm{sech}}^{-1}x$. Find the value of ${\mathrm{sech}}^{-1}\left(\frac{1}{3}\right)$. 

Draw the graph of ${\tanh }^{-1}x$. Find the value of ${\tanh }^{-1}\left(\frac{1}{3}\right)$. 

Draw the graph of ${\tanh }^{-1}x$. Find the value of ${\tanh }^{-1}\left(\frac{1}{2}\right)$. 

Draw the graph of ${\cosh }^{-1}x$. Find the value of ${\cosh }^{-1}\left(7\right)$ in terms of $\log$.

Draw the graph of ${\cosh }^{-1}x$. Find the value of ${\cosh }^{-1}\left(5\right)$ in terms of $\log$.

Draw the graph of ${\cosh }^{-1}x$. Find the value of ${\cosh }^{-1}\left(3\right)$ in terms of $\log$.

Draw the graph of ${\cosh }^{-1}x$. Find the value of ${\cosh }^{-1}\left(2\right)$ in terms of $\log$.

If for $\sinh x=3$, the value of $x$ obtained is in the form of ${\log }_e(a+\sqrt{b})$, then find the value of $a+b$.

If $f\left(x\right)=\cos h^{-1}\left(\frac{1-x}{1+x}\right)$ is well defined, then $f^{\text{'}}\left(x\right)=$

If $\sinh x=\frac{3}{4}$ and $\cosh y=\frac{5}{3}$ then $x+y=$

If $2 \sin h^{-1}\left(\frac{a}{\sqrt{1-a^2}}\right)=\log \left(\frac{1+x}{1-x}\right)$ then $x=$

If ${\mathrm{sech}}^{-1}\left(\frac{1}{2}\right)-{cosech}^{-1}\left(\frac{3}{4}\right)={\log }_{e}k,$ then

If for $\left|x\right|>1$${\tanh }^{-1}\left(\frac{1}{x}\right)+{coth}^{-1}\left(x\right)={\log }_{\mathrm{e}}\left(f\left(x\right)\right),$ then $f\left(-5\right)=$

If $\tan h^{-1}\left(x+iy\right)=\frac{1}{2}\tan h^{-1}\left(\frac{2x}{1+x^2+y^2}\right)+\frac{i}{2}{\tan }^{-1}\left(\frac{2y}{1-x^2-y^2}\right)$, where $x, y\in R$, then $\tan h^{-1}\left(iy\right)=$