What is d/dx(log√((1 + x)/(1 - x))) for -1 < x < 1?

Question 72

MediumOriginal question

What is $\frac{d}{dx}\left(\log\sqrt{\frac{1 + x}{1 - x}}\right)$ for $-1 < x < 1$ ?

\frac{2}{1 - x^{2}}

Correct answerYour answer

-\frac{x}{1 - x^{2}}

Correct answerYour answer

\frac{1}{(1 - x)^{2}}

Correct answerYour answer

\frac{1}{1 - x^{2}}

Correct answerYour answer

Show the answer and explanation

CorrectNot quiteAnswer D

Correct answer: D

$\log\sqrt{\frac{1 + x}{1 - x}} = \frac{1}{2}\left[\log(1 + x) - \log(1 - x)\right]$ .
$\frac{1}{2}\left[\frac{1}{1 + x} + \frac{1}{1 - x}\right] = \frac{1}{2} \cdot \frac{2}{1 - x^{2}} = \frac{1}{1 - x^{2}}$ .

Worth noting

$\frac{2}{1 - x^{2}}$ forgets the $\frac{1}{2}$ that the square root brings down.

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