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

Question 44

HardOriginal question

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

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

Correct answerYour answer

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

Correct answerYour answer

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

Correct answerYour answer

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

Correct answerYour answer

Show the answer and explanation

CorrectNot quiteAnswer C

Correct answer: C

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

Worth noting

$\frac{1}{1 + x^{2}}$ drops the 2 in $2\tan^{-1} x$ .

Ten questions from Chain Rule, Log Differentiation & Rate of Change, at your own pace.