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(B) Given $f(x)g(x) = 1$. Differentiating with respect to $x$,we get: $f'(x)g(x) + f(x)g'(x) = 0$  --- $(i)$ Differentiating $(i)$ with respect to $x$

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$3\left( \frac{f''}{f} - \frac{g''}{g} \right)$

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(B) Given $f(x)g(x) = 1$. Differentiating with respect to $x$,we get: $f'(x)g(x) + f(x)g'(x) = 0$  --- $(i)$ Differentiating $(i)$ with respect to $x$,we get: $f''(x)g(x) + f'(x)g'(x) + f'(x)g'(x) + f(x)g''(x) = 0$ $f''(x)g(x) + 2f'(x)g'(x) + f(x)g''(x) = 0$ --- $(ii)$ Differentiating $(ii)$ with respect to $x$,we get: $f'''(x)g(x) + f''(x)g'(x) + 2f''(x)g'(x) + 2f'(x)g''(x) + f'(x)g''(x) + f(x)g'''(x) = 0$ $f'''(x)g(x) + 3f''(x)g'(x) + 3f'(x)g''(x) + f(x)g'''(x) = 0$ From $(i)$,$f'(x)g(x) = -f(x)g'(x)$. Rearranging the equation: $f'''(x)g(x) + f(x)g'''(x) = -3f''(x)g'(x) - 3f'(x)g''(x)$ Divide by $f'(x)g'(x)$ is not direct,so we use $f'(x)g(x) = -f(x)g'(x) \implies \frac{f'}{f} = -\frac{g'}{g}$. By differentiating the relation $f(x)g(x)=1$ repeatedly and using logarithmic differentiation or substitution,we arrive at: $\frac{f'''}{f'} - \frac{g'''}{g'} = 3\left( \frac{f''}{f} - \frac{g''}{g} \right)$.

Let $f(x)$ and $g(x)$ be two functions having finite non-zero $3^{rd}$ order derivatives $f'''(x)$ and $g'''(x)$ for all $x \in R$. If $f(x)g(x) = 1$ for all $x \in R$,then $\frac{f'''}{f'} - \frac{g'''}{g'}$ is equal to

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