**Problem:**

Differentiate with respect to $x$: $\ln(e^x(\frac{x-1}{x+1})^{\frac{3}{2}})$

**My attempt:**

Let,

$$y=\ln(e^x(\frac{x-1}{x+1})^{\frac{3}{2}})$$

Both $e^x$ and $\frac{x-1}{x+1}$ are positive: $e^x$ can never be negative, and you can take the square root of only positive numbers, so $\frac{x-1}{x+1}$ is positive as well. So, we can apply logarithm properties:

$$=\ln e^x+\ln(\frac{x-1}{x+1})^{\frac{3}{2}}$$

$$=x+\frac{3}{2}\ln(\frac{x-1}{x+1})$$

Now,

$$\frac{dy}{dx}=1+\frac{3}{2}.\frac{x+1}{x-1}.\frac{d}{dx}(\frac{x-1}{x+1})$$

$$=1+\frac{3}{2}.\frac{x+1}{x-1}.\frac{x+1-x+1}{(x+1)^2}$$

$$=1+\frac{3}{2}.\frac{x+1}{x-1}.\frac{2}{(x+1)^2}$$

We can cancel $(x+1)$ and $(x+1)$ in the numerator and denominator because the graph doesn't change.

$$=1+\frac{3}{x^2-1}$$

$$=\frac{x^2+2}{x^2-1}$$

**My book's attempt:**

Let,

$$y=\ln(e^x(\frac{x-1}{x+1})^{\frac{3}{2}})$$

$$=\ln e^x+\ln(\frac{x-1}{x+1})^{\frac{3}{2}}$$

$$=x+\frac{3}{2}\ln(\frac{x-1}{x+1})$$

$$=x+\frac{3}{2}(\ln(x-1)-\ln(x+1))\tag{1}$$

$$...$$

$$\frac{dy}{dx}=\frac{x^2+2}{x^2-1}$$

**Question:**

1. In my book's attempt, is line $(1)$ valid? I think my book's assumption in $(1)$ that $(x-1)$ and $(x+1)$ must be positive is unfounded. I deduced that $\frac{x-1}{x+1}$ is positive; $\frac{x-1}{x+1}$ can be positive even when both $(x-1)$ and $(x+1)$ is negative. So, is line $(1)$ of my book valid?