Find the derivative of $LaTeX: \displaystyle f(x) = \tan^{-1}{\left(x \right)}$ .

Taking the tangent of both sides gives $LaTeX: \displaystyle \tan(y) = x$ . Taking the implicit derivative gives $LaTeX: \displaystyle \sec^2(y)y' = 1$ . Solving for $LaTeX: \displaystyle y'$ gives $LaTeX: \displaystyle y' = \frac{1}{\sec^{2}{\left(y \right)}}$ . Using the trigonometric identity $LaTeX: \displaystyle \tan^2(y)+1 = \sec^2(y)$ gives $LaTeX: \displaystyle y' = \frac{1}{\tan^{2}{\left(y \right)} + 1}$ . Using the fact that $LaTeX: \displaystyle \tan(y)= x$ gives $LaTeX: \displaystyle y' = \frac{1}{x^{2} + 1}$ . Note that the formula for the derivative + the chain rule could have also been used.