If the three functions f(x),g(x),and h(x) are | vedclass

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(C) Given,$h(x) = f(x) \cdot g(x)$ and $f^{\prime}(x) \cdot g^{\prime}(x) = c$.First,find the first derivative of $h(x)$ using the product rule:$h^{\p

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$\frac{h^{\prime \prime}(x)}{h(x)}$

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(C) Given,$h(x) = f(x) \cdot g(x)$ and $f^{\prime}(x) \cdot g^{\prime}(x) = c$.First,find the first derivative of $h(x)$ using the product rule:$h^{\prime}(x) = f^{\prime}(x) \cdot g(x) + f(x) \cdot g^{\prime}(x)$.Next,find the second derivative $h^{\prime \prime}(x)$:$h^{\prime \prime}(x) = \frac{d}{dx}[f^{\prime}(x) \cdot g(x) + f(x) \cdot g^{\prime}(x)]$$h^{\prime \prime}(x) = [f^{\prime \prime}(x) \cdot g(x) + f^{\prime}(x) \cdot g^{\prime}(x)] + [f^{\prime}(x) \cdot g^{\prime}(x) + f(x) \cdot g^{\prime \prime}(x)]$$h^{\prime \prime}(x) = f^{\prime \prime}(x) \cdot g(x) + f(x) \cdot g^{\prime \prime}(x) + 2f^{\prime}(x) \cdot g^{\prime}(x)$.Since $f^{\prime}(x) \cdot g^{\prime}(x) = c$,we have:$h^{\prime \prime}(x) = f^{\prime \prime}(x) \cdot g(x) + f(x) \cdot g^{\prime \prime}(x) + 2c \quad \dots(i)$.Now,evaluate the expression:$\frac{f^{\prime \prime}(x)}{f(x)} + \frac{g^{\prime \prime}(x)}{g(x)} + \frac{2c}{f(x) \cdot g(x)} = \frac{f^{\prime \prime}(x) \cdot g(x) + g^{\prime \prime}(x) \cdot f(x) + 2c}{f(x) \cdot g(x)}$.Substituting from equation $(i)$:$= \frac{h^{\prime \prime}(x)}{h(x)}$.

If the three functions $f(x)$,$g(x)$,and $h(x)$ are such that $h(x) = f(x) \cdot g(x)$ and $f^{\prime}(x) \cdot g^{\prime}(x) = c$,where $c$ is a constant,then $\frac{f^{\prime \prime}(x)}{f(x)} + \frac{g^{\prime \prime}(x)}{g(x)} + \frac{2c}{f(x) \cdot g(x)}$ is equal to:

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