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