Question

If \(f(x) = \sec^{-1} \left(\frac{1}{2x^2 -1}\right)\) and \(g(x) = \tan^{-1} \left(\frac{\sqrt{1 + x^2} - 1}{x}\right)\), then the derivative of \(f(x)\) with respect to \(g(x)\) is?

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

Hint:

Apply chain rule and use derivatives of inverse trigonometric functions carefully.

Answer 1

The Correct Option is D

Use chain rule: \[ \frac{df}{dg} = \frac{\frac{df}{dx}}{\frac{dg}{dx}}. \] Calculate derivatives of \(f(x)\) and \(g(x)\) using inverse secant and inverse tangent differentiation formulas. Simplify ratio to get \[ \frac{df}{dg} = -\frac{4 (1 + x^2)}{\sqrt{1 - x^2}}. \]