#### Question

Let *g* (*x*) be the inverse of an invertible function *f* (*x*) which is derivable at *x* = 3. If *f* (3) = 9 and *f*' (3) = 9, write the value of *g*' (9).

#### Solution

\[\text { We have }, f\left( 3 \right) = 9 , f'\left( 3 \right) = 9\]

\[\text { and g }\left( x \right) = f^{- 1} \left( x \right)\]

\[ \Rightarrow \left( gof \right)\left( x \right) = x\]

\[ \Rightarrow g\left\{ f\left( x \right) \right\} = x\]

\[\Rightarrow \frac{d}{dx}\left[ g\left\{ f\left( x \right) \right\} \right] = 1\]

\[ \Rightarrow g'\left\{ f\left( x \right) \right\}\frac{d}{dx}\left\{ f\left( x \right) \right\} = 1\]

\[ \Rightarrow g'\left\{ f\left( x \right) \right\} \times f'\left( x \right) = 1\]

\[\text { Puting } x = 3, \text { we get }, \]

\[g'\left\{ f\left( 3 \right) \right\} \times f'\left( 3 \right) = 1\]

\[ \Rightarrow g'\left( 9 \right) \times 9 = 1 \left[ \because f\left( 3 \right) = 9 , f'\left( 3 \right) = 9 \right]\]

\[ \Rightarrow g'\left( 9 \right) = \frac{1}{9}\]