You're on a roll — 6. Differentiate $f(x) = \ln (\sinh^{-1} (\tan x))$. Step 1: Apply the chain rule. The outermost function is $\ln(u)$, where $u = \sinh^{-1}(\tan x)$. The derivative of $\ln(u)$ is $\frac{1}{u} \frac{du}{dx}$. $$ f'(x) = \frac{1}{\sinh^{-1}(\tan x)} \cdot \frac{d}{dx}(\sinh^{-1}(\tan x)) $$ Step 2: Differentiate $\sinh^{-1}(v)$, where $v = \tan x$. The derivative of $\sinh^{-1}(v)$ is $\frac{1}{\sqrt{1+v^2}} \frac{dv}{dx}$. $$ \frac{d}{dx}(\sinh^{-1}(\tan x)) = \frac{1}{\sqrt{1+(\tan x)^2}} \cdot \frac{d}{dx}(\tan x) $$ Step 3: Differentiate $\tan x$. The derivative of $\tan x$ is $\sec^2 x$. $$ \frac{d}{dx}(\tan x) = \sec^2 x $$ Step 4: Substitute the derivatives back into the expression for $f'(x)$. $$ f'(x) = \frac{1}{\sinh^{-1}(\tan x)} \cdot \frac{1}{\sqrt{1+\tan^2 x}} \cdot \sec^2 x $$ Step 5: Simplify the expression using the trigonometric identity $1+\tan^2 x = \sec^2 x$. $$ f'(x) = \frac{1}{\sinh^{-1}(\tan x)} \cdot \frac{1}{\sqrt{\sec^2 x}} \cdot \sec^2 x $$ Assuming $\sec x > 0$, then $\sqrt{\sec^2 x} = \sec x$. $$ f'(x) = \frac{1}{\sinh^{-1}(\tan x)} \cdot \frac{1}{\sec x} \cdot \sec^2 x $$ $$ f'(x) = \frac{\sec x}{\sinh^{-1}(\tan x)} $$ The final answer is $\boxed{\frac{\sec x}{\sinh^{-1}(\tan x)}}$. What's next?