Derivative of [e^(-2x)]sqrt(4x-1)

1. Apply the quotient rule, which is:

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

   $f{\left(x \right)} = \sqrt{4 x - 1}$ and $g{\left(x \right)} = e^{2 x}$.

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

   1. Let $u = 4 x - 1$.

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

   3. Then, apply the chain rule. Multiply by $\frac{d}{d x} \left(4 x - 1\right)$:

      1. Differentiate $4 x - 1$ term by term:

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

         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: $4$

         The result is: $4$

      The result of the chain rule is:

      $$\frac{2}{\sqrt{4 x - 1}}$$

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

   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:

      $$2 e^{2 x}$$

   Now plug in to the quotient rule:

   $\left(- 2 \sqrt{4 x - 1} e^{2 x} + \frac{2 e^{2 x}}{\sqrt{4 x - 1}}\right) e^{- 4 x}$

2. Now simplify:

   $$\frac{4 \left(1 - 2 x\right) e^{- 2 x}}{\sqrt{4 x - 1}}$$

The answer is: