Question

If $f(x) = \sin^{-1}\left(\frac{2x}{1 + x^2}\right)$, then $f'\left(\frac{1}{2}\right) =$

• A. $\frac{4}{5}$
• B. $\frac{8}{5}$
• C. $\frac{2}{5}$
• D. 0

Hint:

Expressions like $\frac{2x}{1+x^2}$, $\frac{1-x^2}{1+x^2}$, and $\frac{2x}{1-x^2}$ are classic signatures for the substitution $x = \tan\theta$, transforming them into $\sin 2\theta$, $\cos 2\theta$, and $\tan 2\theta$ respectively.

Answer 1

The Correct Option is B

1. Given the function \(f(x) = \sin^{-1}\left(\frac{2x}{1 + x^2}\right)\), we are tasked to find the derivative \(f'\left(\frac{1}{2}\right)\).
2. To find the derivative, we start by finding \(\frac{d}{dx}\left(\sin^{-1}\left(\frac{2x}{1 + x^2}\right)\right)\).
3. Recall the derivative of \(\sin^{-1}(u)\) is \(\frac{1}{\sqrt{1-u^2}}\cdot \frac{du}{dx}\).
4. Set \(u = \frac{2x}{1 + x^2}\).
5. First, compute the derivative of \(u\):
   \(\frac{du}{dx} = \frac{d}{dx}\left(\frac{2x}{1 + x^2}\right)\).
6. Using the quotient rule, \(\frac{d}{dx}\left(\frac{v}{w}\right) = \frac{v'w - vw'}{w^2}\), where \(v = 2x\) and \(w = 1 + x^2\), we find:
   • \(v' = 2\)
   • \(w' = 2x\)
7. The derivative becomes:
   \(\frac{du}{dx} = \frac{(2)(1 + x^2) - (2x)(2x)}{(1 + x^2)^2}\).
8. Simplify:
   \(\frac{du}{dx} = \frac{2 + 2x^2 - 4x^2}{(1 + x^2)^2} = \frac{2 - 2x^2}{(1 + x^2)^2} = \frac{2(1 - x^2)}{(1 + x^2)^2}\).
9. Now, substitute into the formula for the derivative of \(\sin^{-1}(u)\): 
   \(f'(x) = \frac{1}{\sqrt{1 - \left(\frac{2x}{1 + x^2}\right)^2}} \cdot \frac{2(1 - x^2)}{(1 + x^2)^2}\).
10. Calculate \(\sqrt{1 - \left(\frac{2x}{1 + x^2}\right)^2}\): 
    \(= \sqrt{\frac{(1 + x^2)^2 - 4x^2}{(1 + x^2)^2}}\) 
    \(= \sqrt{\frac{(1 + x^2)^2 - 4x^2}{(1 + x^2)^2}}\) 
    \(= \sqrt{\frac{1 - 2x^2 + x^4}{(1 + x^2)^2}} = \frac{\sqrt{(1-x^2)^2}}{1 + x^2} = \frac{1-x^2}{1 + x^2}\).
11. Finally, the derivative simplifies to: 
    \(f'(x) = \frac{1 + x^2}{1-x^2} \cdot \frac{2(1 - x^2)}{(1 + x^2)^2}\) 
    \(= \frac{2}{1 + x^2}\).
12. Substitute \(x = \frac{1}{2}\) into this derivative: 
    \(f'\left(\frac{1}{2}\right) = \frac{2}{1 + \left(\frac{1}{2}\right)^2}\) 
    \(= \frac{2}{1 + \frac{1}{4}}\) 
    \(= \frac{2}{\frac{5}{4}}\) 
    \(= \frac{2 \cdot 4}{5}\) 
    \(= \frac{8}{5}\).
13. Therefore, the correct answer is \(\frac{8}{5}\).