Let $f (x) = e^{x^{3} + x}$ and $g (x) = f^{- 1} (x)$ then | $g^{''} (1)$ | is

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Solution

$f (x) = e^{x^{3} + x}$
$f^{'} (x) = e^{x^{3} + x} \times (3 x^{2} + 1)$
$∵ g (x) = f^{- 1} (x) \Rightarrow f (g (x)) = x$
$\Rightarrow f^{'} (g (x)) \times g^{'} (x) = 1$
$g^{'} (x) = \frac{1}{f^{'} (g (x))}$
$∵ f (g (x)) = x \Rightarrow f (g (1)) = 1 \Rightarrow g (1) = 0$
$N o w, g^{'} (1) = \frac{1}{f^{'} (g (1))} = \frac{1}{f^{'} (0)} = 1$
$g^{''} (x) = - \frac{1}{(f^{'} (g (x)))^{2}} \times f^{''} (g (x)) \times g^{'} (x)$
$g^{''} (1) = - \frac{1}{(1)^{2}} \times f^{''} (0) \times 1$
$g^{''} (1) = - 1$

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