Answer
\[y'=\sec x\]
Work Step by Step
It is given that\[y=\sinh^{-1} (\tan x)\;\;\;...(1)\]
We will use the formula \[\frac{d}{dx}(\sinh^{-1} x)'=\frac{1}{\sqrt{1+x^2}}\;\;\;...(2)\]
Differentiating (1) with respect to $x$
Using (2)
\[y'=\frac{1}{\sqrt{1+\tan^2 x}}\cdot(\tan x)'\]
\[y'=\frac{1}{\sqrt{1+\tan^2 x}}\cdot(\sec^2 x)\]
\[y'=\frac{1}{\sqrt{\sec^2 x}}\cdot(\sec^2 x)\]
\[\Rightarrow y'=\sec x\]
Hence , \[y'=\sec x\]
