Derivative of (1+x-4sqrtx)/x

1. Apply the quotient rule, which is:

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

   $f{\left(x \right)} = - 4 \sqrt{x} + x + 1$ and $g{\left(x \right)} = x$.

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

   1. Differentiate $- 4 \sqrt{x} + x + 1$ term by term:

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

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

      3. The derivative of a constant times a function is the constant times the derivative of the function.

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

         So, the result is: $- \frac{2}{\sqrt{x}}$

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

   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:

   $\frac{4 \sqrt{x} + x \left(1 - \frac{2}{\sqrt{x}}\right) - x - 1}{x^{2}}$

2. Now simplify:

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

The answer is:

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