**Problem:**

Differentiate with respect to $x$: $\frac{e^{-3x}(3x+5)}{7x-1}$

**My book's attempt:**

Let,

$$y=\frac{e^{-3x}(3x+5)}{7x-1}$$

$$\ln(y)=\ln(\frac{e^{-3x}(3x+5)}{7x-1})\tag{1}$$

$$\text{rest of the math...}$$

**Question:**

1. Isn't taking $\ln()$ unfounded in $(1)$: $\frac{e^{-3x}(3x+5)}{7x-1}$ could a negative number? Don't we have to just use the product and quotient rule to find the derivative here?