Question

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

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

Answer 1

Answer: (C)

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

Answer 2

So that we can write f^–1(y) = g(y)

Now f(x) = x + tan x

Put x = f^–1(y)

f(f^–1 y) = f^–1(y) + tan (f^–1 y)

y = g(y) + tan g(y)

y ⇒ x

x = g(x) + tan g(x) ...(1)

Diff. the function

1 = g'(x) + sec² g(x) . g'(x)

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

from equation (1) tan g(x) = [x – g(x)]

sec² g(x) = 1 + tan² g(x)

= 1 + [x – g(x))²