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 ∈ R

⇒ fog(x) = x for all x ∈ R

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

⇒ f'(g(x)) g'(x) = 1 for all x ∈ R

⇒ f'(g(a)) g'(a) = 1

⇒ 2f'(b) = 1                  [∵ g'(a) = 2 and g(a) = 6]

⇒ f'(b) = $\frac{1}{2}$