If the three function \(f(x), g(x)\) 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 constant, then
\(\frac{f^{\prime \prime}(x)}{f(x)}+\frac{g^{\prime \prime}(x)}{g(x)}+\frac{2 c}{f(x) \cdot g(x)}\)
is equal to

Given, \(h(x)=f(x) \cdot g(x)\) and \(f^{\prime}(x) \cdot g^{\prime}(x)=c\)
Now, \(h^{\prime}(x)=f^{\prime}(x) \cdot g(x)+f(x) \cdot g^{\prime}(x)\)
\(h^{\prime \prime}(x)=f^{\prime \prime}(x) \cdot g(x)+f^{\prime}(x) \cdot g^{\prime}(x)\)
\(+f^{\prime}(x) \cdot g^{\prime}(x)+f(x) \cdot g^{\prime \prime}(x)\)
\(h^{\prime \prime}(x)=f^{\prime \prime}(x) \cdot g(x)+f(x) \cdot g^{\prime \prime}(x)\)
\(+2 f^{\prime}(x) \cdot g^{\prime}(x)\)
\(h^{\prime \prime}(x)=f^{\prime \prime}(x) \cdot g(x)+f(x) \cdot g^{\prime \prime}(x)+2 c \quad \ldots\) (i)
Now, we find
\[
\frac{f^{\prime \prime}(x)}{f(x)}+\frac{g^{\prime \prime}(x)}{g(x)}+\frac{2 c}{f(x) \cdot g(x)}
\]
\[
\begin{aligned}
&=\frac{f^{\prime \prime}(x) \cdot g(x)+g^{\prime \prime}(x) \cdot f(x)+2 c}{f(x) \cdot g(x)} \\
&=\frac{h^{\prime \prime}(x)}{h(x)} \quad[\text { from Eq. (i)] }
\end{aligned}
\]