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?
Show Hint
Apply chain rule and use derivatives of inverse trigonometric functions carefully.
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
\[
\frac{df}{dg} = -\frac{4 (1 + x^2)}{\sqrt{1 - x^2}}.
\]