If f and g are differentiable functions satis | vedclass

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(A) Given that $g(a)=b$,$g^{\prime}(a)=2$,and $f(g(x))=x$ (since $f \circ g = I$).By differentiating both sides of $f(g(x))=x$ with respect to $x$ usi

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$\frac{1}{2}$

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(A) Given that $g(a)=b$,$g^{\prime}(a)=2$,and $f(g(x))=x$ (since $f \circ g = I$).By differentiating both sides of $f(g(x))=x$ with respect to $x$ using the chain rule,we get:$f^{\prime}(g(x)) \cdot g^{\prime}(x) = 1$.Substituting $x=a$ into the equation,we have:$f^{\prime}(g(a)) \cdot g^{\prime}(a) = 1$.Since $g(a)=b$ and $g^{\prime}(a)=2$,the equation becomes:$f^{\prime}(b) \cdot 2 = 1$.Therefore,$f^{\prime}(b) = \frac{1}{2}$.

If $f$ and $g$ are differentiable functions satisfying $g^{\prime}(a)=2$,$g(a)=b$,and $f \circ g = I$,where $I$ is an identity function,then $f^{\prime}(b)$ is equal to

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