Derivative of 2/3x*sqrtx-3x+1

1. Differentiate $\left(\sqrt{x} \frac{2 x}{3} - 3 x\right) + 1$ term by term:

   1. Differentiate $\sqrt{x} \frac{2 x}{3} - 3 x$ term by term:

      1. Apply the quotient rule, which is:

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

         $f{\left(x \right)} = 2 x^{\frac{3}{2}}$ and $g{\left(x \right)} = 3$.

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

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

            1. Apply the power rule: $x^{\frac{3}{2}}$ goes to $\frac{3 \sqrt{x}}{2}$

            So, the result is: $3 \sqrt{x}$

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

         1. The derivative of the constant $3$ is zero.

         Now plug in to the quotient rule:

         $\sqrt{x}$

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

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

         So, the result is: $-3$

      The result is: $\sqrt{x} - 3$

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

   The result is: $\sqrt{x} - 3$

The answer is:

$$\sqrt{x} - 3$$