Differentiating the term, $y=\sqrt{\frac{1-x^2}{1-x}}$ In my calculus book, the above question was given. Since the the term can be simplified, I simplified it to $$\sqrt{(1+x)}$$ which differentiates to $$0.5\sqrt{{\frac{1}{1+x}}}$$ But the answer in the book is $$0.5\sqrt{\frac{1-x}{1-x^2}}$$  I know, if we simplify it gives the same answer.  Then why didn't the author simplify the answer? Is there any special reason?
Hint What if $x=1$ would you do like $\frac{0\times 2}{0}$ and then cancel $0$ .There are indeterminate forms so its safe to keep it like that. Hint you can also make that function continuous by removing singularity at $x=1$ ie(removal of discontinuity at $1$).
Ted's comment is in fact the answer: you didn't differentiate $\sqrt{x}$ correctly. You made it $\frac{1}{2} \sqrt{x}$; it is in fact $\frac{1}{2 \sqrt{x}}$.

Although actually the answer in the book is also incorrect. The author should probably have simplified an expression of the form $\frac{1-x^2}{1-x}$, since the singularity at $x=1$ is removable.