Question

If g is inverse of f and $f^{'}(x) = \frac{1}{1 + x^{n}}$, then $g^{'}(x)$ equals

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

Answer 1

Answer: (C)

$1 + \lbrack g(x)\rbrack^{n}$

Answer 2

Since g is inverse of f. Therefore,

$fog(x) = x$ for all x ⇒ $\frac{d}{dx}\{ fog(x)\} = 1$ for all x

⇒$f^{'}(g(x)).g^{'}(x) = 1$ ⇒ $f^{'}\{ g(x)\} = \frac{1}{g^{'}(x)}$

⇒ $\frac { 1 } { 1 + [ g ( x ) ] ^ { n } } = \frac { 1 } { g ^ { \prime } ( x ) }$ $\left\lbrack \because f^{'}(x) = \frac{1}{1 + x^{n}} \right\rbrack$

⇒ $g^{'}(x) = 1 + \lbrack g(x)\rbrack^{n}$