Problem:
Differentiate with respect to $x$: $\ln\left(e^x\left(\frac{x-1}{x+1}\right)^{\frac{3}{2}}\right)$
My attempt:
Let,
$$y=\ln\left(e^x\left(\frac{x-1}{x+1}\right)^{\frac{3}{2}}\right)$$
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:
$$=\ln e^x+\ln\left(\frac{x-1}{x+1}\right)^{\frac{3}{2}}$$
$$=x+\frac{3}{2}\ln\left(\frac{x-1}{x+1}\right)$$
Now,
$$\frac{dy}{dx}=1+\frac{3}{2}.\frac{x+1}{x-1}.\frac{d}{dx}\left(\frac{x-1}{x+1}\right)$$
$$=1+\frac{3}{2}.\frac{x+1}{x-1}.\frac{x+1-x+1}{(x+1)^2}$$
$$=1+\frac{3}{2}.\frac{x+1}{x-1}.\frac{2}{(x+1)^2}$$
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,
$$y=\ln\left(e^x\left(\frac{x-1}{x+1}\right)^{\frac{3}{2}}\right)$$
$$=\ln e^x+\ln\left(\frac{x-1}{x+1}\right)^{\frac{3}{2}}$$
$$=x+\frac{3}{2}\ln\left(\frac{x-1}{x+1}\right)$$
$$=x+\frac{3}{2}\left(\ln(x-1)-\ln(x+1)\right)\tag{1}$$
$$…$$
$$\frac{dy}{dx}=\frac{x^2+2}{x^2-1}$$
Question:
- 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?
Best Answer
Giving a restriction of the domain is an issue of this problem, because that quantity is defined only for $x<-1\lor x>1$. In my opinion, if this were part of a "gauntlet of exercises" in some textbook, then an indication of the domain should be specified, if the purpose of the exercise lies elsewhere: at the very least, to avoid wasting time. For instance, I think that such a thing would be very important in the chapters about antiderivatives.
You're right: based on the information you've given, $\ln\frac{x-1}{x+1}=\ln(x-1)-\ln(x+1)$ is unwarranted, because the natural way to address the problem would be for all $x$ such that $x<-1\lor x>1$. The book could have solved the issue by saying that $\ln\frac{x-1}{x+1}=\ln\lvert x-1\rvert-\ln\lvert x+1\rvert$ (which is true in the natural domain) and then using $\frac{d}{dt}\ln\lvert t\rvert=\frac1t$.