Question

If f(x) = sin^-1 \((\frac{2x}{1+x^2})\), then f'\((\sqrt3)\) is\(\)

• A. \(\frac{1}{2}\)
• B. \(-\frac{1}{2}\)
• C. \(\frac{1}{\sqrt3}\)
• D. \(-\frac{1}{\sqrt3}\)

Answer 1

The Correct Option is A

If \(f(x) = \sin^{-1}(\frac{2x}{1+x^2})\), then we need to find \(f'(\sqrt{3})\).

We recognize that the expression inside the \(\sin^{-1}\) can be written as \(2\tan \theta / (1+\tan^2 \theta)\), let \(x = \tan \theta\) so \(\theta = \arctan x\)

\(f(x) = \sin^{-1}(\frac{2 \tan \theta}{1+\tan^2 \theta})\)

\(f(x) = \sin^{-1}(\sin 2\theta)\)

\(f(x) = 2 \theta = 2 \arctan x\)

Now we find the derivative \(f'(x)\):

\(f'(x) = 2 \cdot \frac{1}{1+x^2} = \frac{2}{1+x^2}\)

Now we evaluate \(f'(\sqrt{3})\):

\(f'(\sqrt{3}) = \frac{2}{1+(\sqrt{3})^2} = \frac{2}{1+3} = \frac{2}{4} = \frac{1}{2}\)

Therefore, the correct option is (A) \(\frac{1}{2}\).

Answer 2

We have $ f(x) = \sin^{-1}\left(\frac{2x}{1+x^2}\right) $. We can rewrite $ f(x) = 2 \tan^{-1}(x) $.

Then, the derivative is:

$$ f'(x) = \frac{2}{1+x^2} $$

We want to find $ f'(\sqrt{3}) $:

$$ f'(\sqrt{3}) = \frac{2}{1+(\sqrt{3})^2} = \frac{2}{1+3} = \frac{2}{4} = \frac{1}{2} $$