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

1. Apply the quotient rule, which is:

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

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

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

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

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

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

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

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

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

         The result is: $2 x$

      The result of the chain rule is:

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

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

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

   Now plug in to the quotient rule:

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

2. Now simplify:

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

The answer is:

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