Let f and g be differentiable functions satisfying g'(a) = 2, g(a) = b and fog = I(identity function). Then, f'(b) is equal to

$\frac{1}{2}$

We have,

fog = I

⇒  fog(x) = I(x)  for all  x

⇒  f(g(x)) = x  for all  x

⇒  $\frac{d}{d x}\{f(g(x))\}=1$  for all  x

$\Rightarrow \frac{d}{d g(x)}\{f(g(x))\} . \frac{d g(x)}{d x}=1$  for all  x

$\Rightarrow f'(g(x)) g'(x)=1$  for all  x

$\Rightarrow f'(g(a)) g'(a)=1 \Rightarrow f'(b) \times 2=1 \Rightarrow f'(b)=\frac{1}{2}$