Derivative of y=x(sqrt1+x)

1. Apply the product rule:

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

   $f{\left(x \right)} = x$; to find $\frac{d}{d x} f{\left(x \right)}$:

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

   $g{\left(x \right)} = x + \sqrt{1}$; to find $\frac{d}{d x} g{\left(x \right)}$:

   1. Differentiate $x + \sqrt{1}$ term by term:

      1. The derivative of the constant $\sqrt{1}$ is zero.

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

      The result is: $1$

   The result is: $2 x + \sqrt{1}$

2. Now simplify:

   $$2 x + 1$$

The answer is:

$$2 x + 1$$