Answer
${\frac{dy}{dx}}={\frac{1}{1+x^2}}$ where (${\tan{y}=x}$)
Work Step by Step
given that $\tan{y}= {x}$
${\frac{d\tan{y}}{dx}}={\frac{dx}{dx}}$ (on differentiating with respect to x)
and applying the chain rule:
${\frac{d\tan{y}}{dy}}\times{\frac{dy}{dx}}=1$
${\sec^2{y}}{\frac{dy}{dx}}=1$
$({{1+{\tan^2{y}}}}){\frac{dy}{dx}}=1$ (${ 1+{\tan^2{y}}}={\sec^2{y}}$)
$(1+x^2){\frac{dy}{dx}}=1$
${\frac{dy}{dx}}={\frac{1}{1+x^2}}$
