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

Step 1: Understanding the Concept:
We need to find the derivative of an inverse trigonometric function involving a rational algebraic expression, and then evaluate it at a specific point.

Step 2: Key Formula or Approach:
The expression $\frac{2x}{1+x^2}$ strongly suggests the trigonometric substitution $x = \tan\theta$, since $\sin(2\theta) = \frac{2\tan\theta}{1+\tan^2\theta}$. Substitute, simplify the function using inverse trig properties, differentiate the simplified function with respect to $x$, and finally plug in $x = \frac{1}{2}$.

Step 3: Detailed Explanation:
The given function is $f(x) = \sin^{-1}\left(\frac{2x}{1 + x^2}\right)$. Let's use the substitution $x = \tan\theta$. This means $\theta = \tan^{-1}x$. Substitute $x$ into the function: \[ f(x) = \sin^{-1}\left(\frac{2\tan\theta}{1 + \tan^2\theta}\right) \] Using the double angle identity for sine, $\sin 2\theta = \frac{2\tan\theta}{1 + \tan^2\theta}$: \[ f(x) = \sin^{-1}(\sin 2\theta) \] Since we are evaluating the derivative at $x = \frac{1}{2}$ (which is between -1 and 1), $2\theta$ falls within the principal value branch of arcsin $[-\pi/2, \pi/2]$. Therefore, we can simplify directly: \[ f(x) = 2\theta \] Substitute back $\theta = \tan^{-1}x$: \[ f(x) = 2\tan^{-1}x \] Now, differentiate $f(x)$ with respect to $x$: \[ f'(x) = \frac{d}{dx} (2\tan^{-1}x) = 2 \cdot \frac{1}{1 + x^2} = \frac{2}{1 + x^2} \] Finally, evaluate the derivative at $x = \frac{1}{2}$: \[ f'\left(\frac{1}{2}\right) = \frac{2}{1 + \left(\frac{1}{2}\right)^2} \] \[ f'\left(\frac{1}{2}\right) = \frac{2}{1 + \frac{1}{4}} \] \[ f'\left(\frac{1}{2}\right) = \frac{2}{\frac{5}{4}} \] \[ f'\left(\frac{1}{2}\right) = 2 \times \frac{4}{5} = \frac{8}{5} \]

Step 4: Final Answer:
The value of the derivative at $x = 1/2$ is $8/5$.