f and g are two differentiable function which satisfy the condition $g'(a) = 2, g(a) = b$ and $(fog) = I$ (identity function), then $f'(b)$ is equal to

$\frac{1}{2}$

$(fog)x=I(x)=x$ or $f[g(x)]=x$

Differentiating both sides $f'[g(x)]g'(x)=1$

Put x = a, $f'[g(a)]g'(a)=1$

Now, put $g(a) = b$ and $g'(a) = 2$

$f'[b].2=1⇒f'(b)=\frac{1}{2}$