Derivative of log(5*x)*e^(-x)

1. Apply the quotient rule, which is:

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

   $f{\left(x \right)} = \log{\left(5 x \right)}$ and $g{\left(x \right)} = e^{x}$.

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

   1. Let $u = 5 x$.

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

   3. Then, apply the chain rule. Multiply by $\frac{d}{d x} 5 x$:

      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: $5$

      The result of the chain rule is:

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

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

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

   Now plug in to the quotient rule:

   $\left(- e^{x} \log{\left(5 x \right)} + \frac{e^{x}}{x}\right) e^{- 2 x}$

2. Now simplify:

   $$\frac{\left(- x \log{\left(5 x \right)} + 1\right) e^{- x}}{x}$$

The answer is: