Answer
\[\frac{dy}{dx}=\frac{1}{x\sqrt{x^2-1}}\]
Work Step by Step
Let \[y=\sec^{-1} x\]
\[\Rightarrow \sec y=x\;\;\;...(1)\]
Differentiating (1) implicitly with respect to $x$
\[\sec y\tan y\frac{dy}{dx}=1\]
\[\Rightarrow\frac{dy}{dx}=\frac{1}{\sec y\tan y}\]
\[\Rightarrow\frac{dy}{dx}=\frac{1}{\sec y\sqrt{\sec^2 y-1}}\]
Using (1)
\[\Rightarrow\frac{dy}{dx}=\frac{1}{x\sqrt{x^2-1}}\]
Hence \[\Rightarrow\frac{dy}{dx}=\frac{1}{x\sqrt{x^2-1}}\]
