Derivative of y=(e^x+1)log2x

1. Apply the product rule:

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

   $f{\left(x \right)} = e^{x} + 1$; to find $\frac{d}{d x} f{\left(x \right)}$:

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

      1. The derivative of $e^{x}$ is itself.

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

      The result is: $e^{x}$

   $g{\left(x \right)} = \log{\left(2 x \right)}$; to find $\frac{d}{d x} g{\left(x \right)}$:

   1. Let $u = 2 x$.

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

   3. Then, apply the chain rule. Multiply by $\frac{d}{d x} 2 x$:

      1. 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: $2$

      The result of the chain rule is:

      $$\frac{1}{x}$$

   The result is: $e^{x} \log{\left(2 x \right)} + \frac{e^{x} + 1}{x}$

2. Now simplify:

   $$\frac{x e^{x} \log{\left(2 x \right)} + e^{x} + 1}{x}$$

The answer is:

$$\frac{x e^{x} \log{\left(2 x \right)} + e^{x} + 1}{x}$$