If g is the inverse function of $f and f^{'} (x) = \sin x then g^{'} (x)$ is

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$\cos e c (g (x))$

sin(g(x))

$- \frac{1}{\sin (g (x))}$

$\cos (g (x))$

answer is A.

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Detailed Solution

$g (x) = f^{- 1} (x) \Rightarrow f (g (x)) = x \Rightarrow f^{'} (g (x)) \cdot g^{'} (x) = 1 \Rightarrow \sin (g (x)) . g^{'} (x) = 1 \Rightarrow g^{'} (x) = \cos e c (g (x))$

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