Let f and g be twice differentiable functions | vedclass

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(D) Given $f(x) \cdot g(x) = 1$. Differentiating both sides with respect to $x$,we get: $f(x) \cdot g'(x) + g(x) \cdot f'(x) = 0$  --- $(1)$ Different

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$2\left(\frac{f'(x)}{f(x)}\right)^2$

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(D) Given $f(x) \cdot g(x) = 1$. Differentiating both sides with respect to $x$,we get: $f(x) \cdot g'(x) + g(x) \cdot f'(x) = 0$  --- $(1)$ Differentiating equation $(1)$ again with respect to $x$: $f(x) \cdot g''(x) + f'(x) \cdot g'(x) + g'(x) \cdot f'(x) + g(x) \cdot f''(x) = 0$ $f(x) \cdot g''(x) + g(x) \cdot f''(x) + 2f'(x) \cdot g'(x) = 0$ Dividing the entire equation by $f(x) \cdot g(x)$ (which is $1$): $\frac{f(x) \cdot g''(x)}{f(x) \cdot g(x)} + \frac{g(x) \cdot f''(x)}{f(x) \cdot g(x)} + \frac{2f'(x) \cdot g'(x)}{f(x) \cdot g(x)} = 0$ $\frac{g''(x)}{g(x)} + \frac{f''(x)}{f(x)} + 2 \cdot \frac{f'(x)}{f(x)} \cdot \frac{g'(x)}{g(x)} = 0$ --- $(2)$ From equation $(1)$,we have $f(x) \cdot g'(x) = -g(x) \cdot f'(x)$,which implies $\frac{g'(x)}{g(x)} = -\frac{f'(x)}{f(x)}$. Substituting this into equation $(2)$: $\frac{f''(x)}{f(x)} + \frac{g''(x)}{g(x)} + 2 \cdot \frac{f'(x)}{f(x)} \cdot \left(-\frac{f'(x)}{f(x)}\right) = 0$ $\frac{f''(x)}{f(x)} + \frac{g''(x)}{g(x)} - 2 \left(\frac{f'(x)}{f(x)}\right)^2 = 0$ $\frac{f''(x)}{f(x)} + \frac{g''(x)}{g(x)} = 2 \left(\frac{f'(x)}{f(x)}\right)^2$.

Let $f$ and $g$ be twice differentiable functions such that $f(x) \cdot g(x) = 1$ for all $x \in R$ and $f'$ and $g'$ are never zero. Then $\frac{f''(x)}{f(x)} + \frac{g''(x)}{g(x)}$ equals:

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