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प्रश्न
If \[f\left( x \right) = \log \left\{ \frac{u \left( x \right)}{v \left( x \right)} \right\}, u \left( 1 \right) = v \left( 1 \right) \text{ and }u' \left( 1 \right) = v' \left( 1 \right) = 2\] , then find the value of `f' (1)` ?
उत्तर
\[\text{ We have, }f\left( x \right) = \log\left\{ \frac{u\left( x \right)}{v\left( x \right)} \right\}\]
\[\text{ and,} \]
\[ u\left( 1 \right) = v\left( 1 \right) , u'\left( 1 \right) = v'\left( 1 \right) = 2 ....... \left( \text{i} \right)\]
\[\Rightarrow f'\left( x \right) = \frac{d}{dx}\left[ \log\left\{ \frac{u\left( x \right)}{v\left( x \right)} \right\} \right]\]
\[ \Rightarrow f'\left( x \right) = \frac{1}{\left[ \frac{u\left( x \right)}{v\left( x \right)} \right]} \times \frac{d}{dx}\left[ \frac{u\left( x \right)}{v\left( x \right)} \right]\]
\[ \Rightarrow f'\left( x \right) = \frac{v\left( x \right)}{u\left( x \right)} \times \left[ \frac{v\left( x \right)\frac{d}{dx}\left\{ u\left( x \right) \right\} - u\left( x \right)\frac{d}{dx}\left\{ v\left( x \right) \right\}}{\left\{ v\left( x \right) \right\}^2} \right] \]
\[ \Rightarrow f'\left( x \right) = \frac{v\left( x \right)}{u\left( x \right)} \times \left[ \frac{v\left( x \right) \times u'\left( x \right) - u\left( x \right) \times v'\left( x \right)}{\left\{ v\left( x \right) \right\}^2} \right]\]
\[\text{ Putting x = 1, we get }, \]
\[f'\left( 1 \right) = \frac{v\left( 1 \right)}{u\left( 1 \right)} \times \left[ \frac{v\left( 1 \right) \times u'\left( 1 \right) - u\left( 1 \right) \times v'\left( 1 \right)}{\left\{ v\left( 1 \right) \right\}^2} \right]\]
\[ \Rightarrow f'\left( 1 \right) = 1 \times \left[ \frac{u\left( 1 \right) \times 2 - u\left( 1 \right) \times 2}{\left\{ u\left( 1 \right) \right\}^2} \right] .........\left[ \text{ Using eqn } \left( \text{i} \right) \right]\]
\[ \Rightarrow f'\left( 1 \right) = \left[ \frac{0}{\left\{ u\left( 1 \right) \right\}^2} \right] \]
\[ \Rightarrow f'\left( 1 \right) = 0\]
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