What is dy/dx for y = cos^-1((1 - x^2)/(1 + x^2)), 0 < x < 1?

Question 45

HardOriginal question

What is $\frac{dy}{dx}$ for $y = \cos^{-1}\left(\frac{1 - x^{2}}{1 + x^{2}}\right)$ , $0 < x < 1$ ?

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

Correct answerYour answer

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

Correct answerYour answer

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

Correct answerYour answer

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

Correct answerYour answer

Show the answer and explanation

CorrectNot quiteAnswer D

Correct answer: D

Put $x = \tan \theta$ : $\frac{1 - \tan^{2} \theta}{1 + \tan^{2} \theta} = \cos 2\theta$ , so the function is $2\theta = 2\tan^{-1} x$ .
So $\frac{dy}{dx} = \frac{2}{x^{2} + 1}$ .

Worth noting

$-\frac{2}{1 + x^{2}}$ keeps the minus sign of $\cos^{-1}$ , but after the substitution the function is $2\tan^{-1} x$ .

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