Derivative of e^(2x)cosx

1. Apply the product rule:

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

   $f{\left(x \right)} = e^{2 x}$; to find $\frac{d}{d x} f{\left(x \right)}$:

   1. Let $u = 2 x$.

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

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

      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:

      $$2 e^{2 x}$$

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

   1. The derivative of cosine is negative sine:

      $$\frac{d}{d x} \cos{\left(x \right)} = - \sin{\left(x \right)}$$

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

2. Now simplify:

   $$\left(- \sin{\left(x \right)} + 2 \cos{\left(x \right)}\right) e^{2 x}$$

The answer is:

$$\left(- \sin{\left(x \right)} + 2 \cos{\left(x \right)}\right) e^{2 x}$$