Answer
${dy}=(\frac{2x}{1+x^2}){dx}$
Work Step by Step
We evaluate the function: $y=\ln{(1+x^2)}$
on differentiating the above:
$\frac{dy}{dx}=\frac{d{(\ln{(1+x^2)}})}{dx}$
$\frac{dy}{dx}=\frac{1}{(1+x^2)}\frac{d(1+x^2)}{dx}$
or $\frac{dy}{dx}=\frac{2x}{1+x^2}$
or ${dy}=(\frac{2x}{1+x^2}){dx}$
The final answer is: ${dy}=(\frac{2x}{1+x^2}){dx}$
