If ` y= sin^(-1)"" 2x . ""sqrt((1-x^2))+ sec^(-1)"" (1)/( sqrt((1-x^2))`, then `dy// dx` =

Options

A. `-(1)/( sqrt(1-x^2))`
B. `(2)/(sqrt(1-x^2))`
C. `(3)/( sqrt((1-x^2)))`
D. None

Answer

To find the derivative \( \frac{dy}{dx} \) for the function

\[ y = \sin^{-1}(2x \sqrt{1-x^2}) + \sec^{-1}\left(\frac{1}{\sqrt{1-x^2}}\right), \]

we will use substitution and differentiation techniques.

### Step-by-Step Solution:

1. **Substitution**: Let \( x = \sin \theta \). Then, \( dx = \cos \theta \, d\theta \) and \( \sqrt{1-x^2} = \sqrt{1-\sin^2 \theta} = \cos \theta \).

2. **Rewrite \( y \)**: Substitute \( x \) in the equation:
\[ y = \sin^{-1}(2 \sin \theta \cos \theta) + \sec^{-1}\left(\frac{1}{\cos \theta}\right). \]
Using the double angle identity, \( 2 \sin \theta \cos \theta = \sin(2\theta) \), we can simplify:
\[ y = \sin^{-1}(\sin(2\theta)) + \sec^{-1}(\sec \theta). \]
Since \( \sin^{-1}(\sin(2\theta)) = 2\theta \) (for \( \theta \) in the principal range) and \( \sec^{-1}(\sec \theta) = \theta \), we have:
\[ y = 2\theta + \theta = 3\theta. \]

3. **Express \( \theta \)**: Since \( x = \sin \theta \), we find \( \theta = \sin^{-1}(x) \). Thus,
\[ y = 3 \sin^{-1}(x). \]

4. **Differentiate \( y \)**: Now we differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = 3 \cdot \frac{d}{dx}(\sin^{-1}(x)). \]
The derivative of \( \sin^{-1}(x) \) is given by:
\[ \frac{d}{dx}(\sin^{-1}(x)) = \frac{1}{\sqrt{1-x^2}}. \]
Therefore,
\[ \frac{dy}{dx} = 3 \cdot \frac{1}{\sqrt{1-x^2}}. \]

5. **Final Result**: Thus, the derivative is:
\[ \frac{dy}{dx} = \frac{3}{\sqrt{1-x^2}}. \]