If f(x) = ⁻¹ ( 3x - x^3 1 - 3x^2 ) and g(x) = ⁻¹ ( 1 - x^2 1 + x^2 ) , then _ x a f(x) - f(a) g(x) - g(a) , (0 < a < 1 2 ) is

Step-by-step solution

We need to evaluate _ x a f(x) - f(a) g(x) - g(a) . This is the definition of the derivative of f(x) with respect to g(x) at x = a , which is equal to f'(a) g'(a) . Given f(x) = ⁻¹ ( 3x - x^3 1 - 3x^2 ) . Let x = . Since 0 3x - x^3 1 - 3x^2 = 3 - ^3 1 - 3 ^2 = 3 Since 0 0 . f(x) = ⁻¹( 3 ) = ⁻¹ ( ( 2 - 3 ) ) Since 2 - 3 (0, 2 ) , it lies in the principal range of ⁻¹ . f(x) = 2 - 3 = 2 - 3 ⁻¹x Differentiating with respect to x : f'(x) = -3 1 + x^2 Now, g(x) = ⁻¹ ( 1 - x^2 1 + x^2 ) . Substitute x = again: 1 - x^2 1 +