Question

If f(x) = x + tan x and g(x) is the inverse of f(x), then g'(x) is equal to

• A. $\frac{1}{1 + \lbrack g(x) - x\rbrack^{2}}$
• B. $\frac{1}{2 + \lbrack g(x) + x\rbrack^{2}}$
• C. $\frac{1}{2 + \lbrack g(x) - x\rbrack^{2}}$
• D. None of these

Answer 1

Answer: (C)

$\frac{1}{2 + \lbrack g(x) - x\rbrack^{2}}$

Answer 2

Given, f(x) = x + tan x

̃ f(f^–1(x)) = f^–1(x) + tan (f^–1 (x))

̃ x = g(x) + tan(g(x)) …(i)

[Q g(x) = f^–1 (x)]

̃ 1 = g'(x) + sec²(g(x))g'(x)

̃ g'(x) = $\frac{1}{1 + \sec^{2}(g(x))}$

̃ g'(x) = $\frac{1}{2 + \tan^{2}(g(x))}$

̃ g'(x) = $\frac{1}{2 + \lbrack x - g(x))\rbrack^{2}}$ [from Eq. (i)]