Derivative of y/sqrt(1+y^2)

1. Apply the quotient rule, which is:

   $$\frac{d}{d y} \frac{f{\left(y \right)}}{g{\left(y \right)}} = \frac{- f{\left(y \right)} \frac{d}{d y} g{\left(y \right)} + g{\left(y \right)} \frac{d}{d y} f{\left(y \right)}}{g^{2}{\left(y \right)}}$$

   $f{\left(y \right)} = y$ and $g{\left(y \right)} = \sqrt{y^{2} + 1}$.

   To find $\frac{d}{d y} f{\left(y \right)}$:

   1. Apply the power rule: $y$ goes to $1$

   To find $\frac{d}{d y} g{\left(y \right)}$:

   1. Let $u = y^{2} + 1$.

   2. Apply the power rule: $\sqrt{u}$ goes to $\frac{1}{2 \sqrt{u}}$

   3. Then, apply the chain rule. Multiply by $\frac{d}{d y} \left(y^{2} + 1\right)$:

      1. Differentiate $y^{2} + 1$ term by term:

         1. The derivative of the constant $1$ is zero.

         2. Apply the power rule: $y^{2}$ goes to $2 y$

         The result is: $2 y$

      The result of the chain rule is:

      $$\frac{y}{\sqrt{y^{2} + 1}}$$

   Now plug in to the quotient rule:

   $\frac{- \frac{y^{2}}{\sqrt{y^{2} + 1}} + \sqrt{y^{2} + 1}}{y^{2} + 1}$

2. Now simplify:

   $$\frac{1}{\left(y^{2} + 1\right)^{\frac{3}{2}}}$$

The answer is: