Apply the chain rule. The outermost function is (u), where u = ^-1( x).

The outermost function is $\ln(u)$, where $u = \sinh^{-1}(\tan x)$.
The derivative of $\ln(u)$ is $\frac{1}{u} \frac{du}{dx}$.

$$f'(x) = \frac{1}{\sinh^{-1}(\tan x)} \cdot \frac{d}{dx}(\sinh^{-1}(\tan x))$$

The derivative of $\sinh^{-1}(v)$ is $\frac{1}{\sqrt{1+v^2}} \frac{dv}{dx}$.

$$\frac{d}{dx}(\sinh^{-1}(\tan x)) = \frac{1}{\sqrt{1+(\tan x)^2}} \cdot \frac{d}{dx}(\tan x)$$

The derivative of $\tan x$ is $\sec^2 x$.

$$\frac{d}{dx}(\tan x) = \sec^2 x$$

$$f'(x) = \frac{1}{\sinh^{-1}(\tan x)} \cdot \frac{1}{\sqrt{\sec^2 x}} \cdot \sec^2 x$$

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 $\frac{\sec x}{\sinh^{-1}(\tan x)}$.