Derivative of y=lnx²+2x-e^(-2x)

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

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

      1. Let $u = \log{\left(x \right)}$.

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

      3. Then, apply the chain rule. Multiply by $\frac{d}{d x} \log{\left(x \right)}$:

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

         The result of the chain rule is:

         $$\frac{2 \log{\left(x \right)}}{x}$$

      4. 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 is: $2 + \frac{2 \log{\left(x \right)}}{x}$

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

      1. Let $u = - 2 x$.

      2. The derivative of $e^{u}$ is itself.

      3. Then, apply the chain rule. Multiply by $\frac{d}{d x} \left(- 2 x\right)$:

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

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

            So, the result is: $-2$

         The result of the chain rule is:

         $$- 2 e^{- 2 x}$$

      So, the result is: $2 e^{- 2 x}$

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

The answer is:

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