Question

\(\frac{d}{dx} \left( \sin^{-1} 2x \right)\)

• A. \( \frac{1}{\sqrt{1 - 4x^2}} \)
• B. \( \frac{2}{\sqrt{1 - x^2}} \)
• C. \( \frac{2}{\sqrt{1 - 4x^2}} \)
• D. \( \frac{n}{2} - \cos^{-1} 2x \)

Hint:

For inverse trigonometric functions, always apply the chain rule when there is a function inside the inverse function.

Answer 1

The Correct Option is C

We are asked to differentiate \( \sin^{-1} 2x \), which is the inverse sine of \( 2x \). Step 1: Apply the derivative of inverse sine
The derivative of \( \sin^{-1}(u) \) with respect to \( u \) is: \[ \frac{d}{du} \sin^{-1}(u) = \frac{1}{\sqrt{1 - u^2}} \] For \( \sin^{-1}(2x) \), we apply the chain rule. Let \( u = 2x \), so: \[ \frac{d}{dx} \sin^{-1}(2x) = \frac{2}{\sqrt{1 - (2x)^2}} = \frac{2}{\sqrt{1 - 4x^2}} \]