Question

If f (x) = log_x² (log x), the `f' (x)` at x = e is ____________ .

Options

• 0

• 1

• 1/e

• 1/2e

Answer 1

1/2 e

\[\text{ We have,} f\left( x \right) = \log_{x^2} \left( \log x \right)\]
\[ \Rightarrow f\left( x \right) = \frac{\log\left( \log x \right)}{\log x^2} \]
\[ \Rightarrow f\left( x \right) = \frac{\log\left( \log x \right)}{2 \log x}\]
\[ \Rightarrow f'\left( x \right) = \frac{1}{2} \times \frac{d}{dx}\left\{ \frac{\log\left( \log x \right)}{\log x} \right\}\]
\[ \Rightarrow f'\left( x \right) = \frac{1}{2} \times \left\{ \frac{\frac{1}{\log x} \times \frac{1}{x} \times \log x - \frac{\log\left( \log x \right)}{x}}{\left( \log x \right)^2} \right\}\]
\[ \Rightarrow f'\left( x \right) = \frac{1}{2} \times \left\{ \frac{\frac{1}{x} - \frac{\log\left( \log x \right)}{x}}{\left( \log x \right)^2} \right\}\]
\[ \Rightarrow f'\left( e \right) = \frac{1}{2} \times \left\{ \frac{\frac{1}{e} - \frac{\log\left( \log e \right)}{e}}{\left( \log e \right)^2} \right\} \left[ \text{ Putting x } = e \right]\]
\[ \Rightarrow f'\left( e \right) = \frac{1}{2} \times \left\{ \frac{\frac{1}{e}}{1} \right\}\]
\[ \Rightarrow f'\left( e \right) = \frac{1}{2e}\]