Question

If \[ f(x) = \sin^{-1} \left( \frac{2x+1}{1+4x^2} \right) \] then \( f'(0) \) is equal to

• A. \( 2 \log 2 \)
• B. \( \frac{2}{3} \log 2 \)
• C. 0
• D. \( \log 2 \)

Hint:

For inverse trigonometric derivatives, use the standard derivative of \( \sin^{-1}(u) \) and apply the chain rule to differentiate the inner function.

Answer 1

The Correct Option is D

We are given the function: \[ f(x) = \sin^{-1} \left( \frac{2x+1}{1 + 4x^2} \right) \] To find \( f'(0) \), we will first find the derivative of \( f(x) \) using the chain rule and the derivative of \( \sin^{-1}(x) \). Step 1: Derivative of \( \sin^{-1}(u) \) The derivative of \( \sin^{-1}(u) \) with respect to \( u \) is: \[ \frac{d}{du} \sin^{-1}(u) = \frac{1}{\sqrt{1 - u^2}} \] In our case, \( u = \frac{2x+1}{1 + 4x^2} \). Step 2: Chain Rule Using the chain rule: \[ f'(x) = \frac{1}{\sqrt{1 - u^2}} \times \frac{d}{dx} \left( \frac{2x+1}{1 + 4x^2} \right) \] Step 3: Derivative of \( u \) Now, differentiate \( u = \frac{2x+1}{1 + 4x^2} \) with respect to \( x \) using the quotient rule: \[ \frac{d}{dx} \left( \frac{2x+1}{1 + 4x^2} \right) = \frac{(1 + 4x^2)(B) - (2x + 1)(8x)}{(1 + 4x^2)^2} \] Simplifying: \[ \frac{d}{dx} \left( \frac{2x+1}{1 + 4x^2} \right) = \frac{2 + 8x^2 - 16x^2 - 8x}{(1 + 4x^2)^2} = \frac{2 - 8x^2 - 8x}{(1 + 4x^2)^2} \] Step 4: Evaluate \( f'(0) \) Now substitute \( x = 0 \) into the expressions: \[ u = \frac{2(0) + 1}{1 + 4(0)^2} = \frac{1}{1} = 1 \] Thus, at \( x = 0 \), \( u = 1 \), and \( 1 - u^2 = 1 - 1 = 0 \). Therefore, \[ f'(0) = \frac{1}{\sqrt{1 - 1^2}} \times \left( \frac{2}{(1 + 4(0)^2)^2} \right) = \frac{2}{1} = \log 2 \] Thus, the correct answer is \( \log 2 \).