Derivative of y=logx(3x^2-x+5)

1. Apply the product rule:

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

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

   1. The derivative of $\log{\left(x \right)}$ is $\frac{1}{x}$.

   $g{\left(x \right)} = \left(3 x^{2} - x\right) + 5$; to find $\frac{d}{d x} g{\left(x \right)}$:

   1. Differentiate $\left(3 x^{2} - x\right) + 5$ term by term:

      1. Differentiate $3 x^{2} - x$ term by term:

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

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

            So, the result is: $6 x$

         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: $-1$

         The result is: $6 x - 1$

      2. The derivative of the constant $5$ is zero.

      The result is: $6 x - 1$

   The result is: $\left(6 x - 1\right) \log{\left(x \right)} + \frac{\left(3 x^{2} - x\right) + 5}{x}$

2. Now simplify:

   $$6 x \log{\left(x \right)} + 3 x - \log{\left(x \right)} - 1 + \frac{5}{x}$$

The answer is:

$$6 x \log{\left(x \right)} + 3 x - \log{\left(x \right)} - 1 + \frac{5}{x}$$