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f and g are twice differentiable functions and f^{\prime} \neq 0, g^{\prime} \neq 0. If f(x) g(x)=1, then \frac{f^{\prime \prime}(x)}{f^{\prime}(x)}-\frac{g^{\prime}(x)}{g^{\prime}(x)} is equal to


 

Option: 1

\frac{-2 g^{\prime}(x)}{g(x)}


Option: 2

\frac{g^{\prime}(x)}{g(x)}


Option: 3

\frac{2 f^{\prime}(x)}{f(x)}


Option: 4

\frac{-f^{\prime}(x)}{f(x)}


Answers (1)

best_answer

Differentiating f \ g=1, we get f^{\prime} g+g^{\prime} f=0 \Rightarrow \frac{f^{\prime}}{f}=\frac{g^{\prime}}{g} Differentiating again, we get
 \begin{array}{lc} f^{\prime} g+2 f^{\prime} g^{\prime}+g^{\prime \prime} f=0 \\ \Rightarrow & \frac{f^{\prime \prime}}{f^{\prime}} \cdot \frac{g}{g^{\prime}}+2+\frac{g^{\prime \prime}}{g^{\prime}} \cdot \frac{f}{f^{\prime}}=0 \\ \Rightarrow & \frac{f^{\prime \prime}}{f^{\prime}} \cdot \frac{g}{g^{\prime}}+2-\frac{g^{\prime \prime}}{g^{\prime}} \cdot \frac{g}{g^{\prime}}=0 \\ \Rightarrow & \frac{f^{\prime \prime}}{f^{\prime}}-\frac{g^{\prime \prime}}{g^{\prime}}=-\frac{2 g^{\prime}}{g} \\ \text { or } & \frac{f^{\prime \prime}}{f^{\prime}}\left(-\frac{f}{f^{\prime}}\right)+2+\frac{g^{\prime \prime}}{g^{\prime}} \cdot \frac{f}{f^{\prime}}=0 \\ \Rightarrow & \frac{f^{\prime \prime}}{f^{\prime}}-\frac{g^{\prime \prime}}{g^{\prime}}=\frac{2 f^{\prime}}{f} \end{array}

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Divya Prakash Singh

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