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(D) Given that $g(x) = f^{-1}(x)$.By the definition of inverse functions, we have $f(g(x)) = x$.Differentiating both sides with respect to $x$ using t

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$\frac{4}{3}$

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(D) Given that $g(x) = f^{-1}(x)$.By the definition of inverse functions, we have $f(g(x)) = x$.Differentiating both sides with respect to $x$ using the chain rule, we get $f^{\prime}(g(x)) \cdot g^{\prime}(x) = 1$.Therefore, $f^{\prime}(g(x)) = \frac{1}{g^{\prime}(x)}$.We are given $g(5) = 6$ and $g^{\prime}(5) = \frac{3}{4}$.Substituting $x = 5$ into the equation $f^{\prime}(g(x)) = \frac{1}{g^{\prime}(x)}$, we get $f^{\prime}(g(5)) = \frac{1}{g^{\prime}(5)}$.Since $g(5) = 6$, this becomes $f^{\prime}(6) = \frac{1}{g^{\prime}(5)}$.Substituting the value $g^{\prime}(5) = \frac{3}{4}$, we get $f^{\prime}(6) = \frac{1}{3/4} = \frac{4}{3}$.

Let $f$ and $g$ be two differentiable functions satisfying $g^{\prime}(5)=\frac{3}{4}$, $g(5)=6$ and $g=f^{-1}$. Then $f^{\prime}(6)$ is equal to

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