Question

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

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

Answer 1

Answer: (C)

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

Answer 2

Q g(x) = f^–1 (x) Ž f (g(x)) = x differentiable

Ž f '(g(x)) . g'(x) = 1

Ž g'(x) = $\frac{1}{f'(g(x))}$ ….(1)

Now from curve

f '(x) = 1 + sec²x

f '(g(x)) = 1 + sec²g(x) = 2 + tan²g(x) …(2)

Now from curve

f (g(x)) = g(x) + tan g(x)

x = g(x) + tan g(x)

Ž tan g(x) = x – g(x) …(3)

from (1), (2), (3)

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