Question

The derivative of \(\sin^{-1}(2x^2 - 1)\) with respect to \(\sin^{-1}x\) is:

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

Hint:

When finding the derivative of a function with respect to another function, use the chain rule: \[ \frac{dy}{dz} = \frac{dy/dx}{dz/dx} \] Also, always simplify radicals carefully when inverse trigonometric functions are involved.

Answer 1

The Correct Option is C

Let: \[ y = \sin^{-1}(2x^2 - 1), \quad z = \sin^{-1}x \] We are asked to find: \[ \frac{dy}{dz} = \frac{dy/dx}{dz/dx} \] Differentiate \(y = \sin^{-1}(2x^2 - 1)\) with respect to \(x\): \[ \frac{dy}{dx} = \frac{1}{\sqrt{1 - (2x^2 - 1)^2}} \cdot \frac{d}{dx}(2x^2 - 1) = \frac{1}{\sqrt{1 - (4x^4 - 4x^2 + 1)}} \cdot 4x \] Simplifying the expression inside the square root: \[ 1 - (4x^4 - 4x^2 + 1) = -4x^4 + 4x^2 = 4x^2(1 - x^2) \] Therefore: \[ \frac{dy}{dx} = \frac{4x}{\sqrt{4x^2(1 - x^2)}} = \frac{4x}{2x\sqrt{1 - x^2}} = \frac{2}{\sqrt{1 - x^2}} \] Now differentiate \(z = \sin^{-1}x\) with respect to \(x\): \[ \frac{dz}{dx} = \frac{1}{\sqrt{1 - x^2}} \] Now compute: \[ \frac{dy}{dz} = \frac{dy/dx}{dz/dx} = \frac{\dfrac{2}{\sqrt{1 - x^2}}}{\dfrac{1}{\sqrt{1 - x^2}}} = 2 \]