Let f and g be differentiable functions on R | vedclass

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(C) Given that $f \circ g$ is the identity function, we have $f(g(x)) = x$ for all $x \in R$.Differentiating both sides with respect to $x$ using the

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$1/5$

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(C) Given that $f \circ g$ is the identity function, we have $f(g(x)) = x$ for all $x \in R$.Differentiating both sides with respect to $x$ using the chain rule, we get:$f^{\prime}(g(x)) \cdot g^{\prime}(x) = 1$.We are given that $g(a) = b$ and $g^{\prime}(a) = 5$.Substituting $x = a$ into the differentiated equation:$f^{\prime}(g(a)) \cdot g^{\prime}(a) = 1$.Substituting the known values $g(a) = b$ and $g^{\prime}(a) = 5$:$f^{\prime}(b) \cdot 5 = 1$.Therefore, $f^{\prime}(b) = 1/5$.

Let $f$ and $g$ be differentiable functions on $R$ such that $f \circ g$ is the identity function. If for some $a, b \in R$, $g^{\prime}(a) = 5$ and $g(a) = b$, then $f^{\prime}(b)$ is equal to:

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