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(A) Given that $f \circ g = I$,where $I$ is the identity function,we have $(f \circ g)(x) = x$ for all $x$ in the domain. Applying the chain rule to d

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$1/2$

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(A) Given that $f \circ g = I$,where $I$ is the identity function,we have $(f \circ g)(x) = x$ for all $x$ in the domain. Applying the chain rule to differentiate both sides with respect to $x$,we get $f'(g(x)) \cdot g'(x) = 1$. Substituting $x = a$ into the equation,we obtain $f'(g(a)) \cdot g'(a) = 1$. Given $g(a) = b$ and $g'(a) = 2$,we substitute these values into the equation: $f'(b) \cdot 2 = 1$. Therefore,$f'(b) = 1/2$.

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

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