Q82 / 180 · KCET 2009 · Mathematics

If $f(x)=\log _{x^{2}}\left(\log _{e} x\right)$, then $f^{\prime}(x)$ at $x=e$ is

  • A
    1
  • B
    $\frac{1}{\mathrm{e}}$
  • Correct answer
    C
    $\frac{1}{2 \mathrm{e}}$
  • D
    0

Solution

Given, $\mathrm{f}(\mathrm{x})=\log _{\mathrm{x}^{2}}\left(\log _{\mathrm{e}} \mathrm{x}\right)=\frac{1}{2} \log _{\mathrm{x}}\left(\log _{\mathrm{e}} \mathrm{x}\right)$ $\Rightarrow \quad \mathrm{f}(\mathrm{x})=\frac{1}{2} \frac{\log _{\mathrm{e}} \log _{\mathrm{e}} \mathrm{x}}{\log _{\mathrm{e}} \mathrm{x}}$ $\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{2} \frac{\log _{\mathrm{e}} \mathrm{x}\left[\frac{1}{\mathrm{x} \log _{\mathrm{e}} \mathrm{x}}\right]-\log _{\mathrm{e}} \log _{\mathrm{e}} \mathrm{x} \times \frac{1}{\mathrm{x}}}{\left(\log _{\mathrm{e}} \mathrm{x}\right)^{2}}$ $\Rightarrow \quad \mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{2} \frac{\frac{1}{\mathrm{x}}-\frac{1}{\mathrm{x}} \log _{\mathrm{e}} \log _{\mathrm{e}} \mathrm{x}}{\left(\log _{\mathrm{e}} \mathrm{x}\right)^{2}}$ At $\mathrm{x}=\mathrm{e}, \mathrm{f}^{\prime}(\mathrm{e})=\frac{1}{2} \frac{\frac{1}{\mathrm{e}}-\frac{1}{\mathrm{e}} \log _{\mathrm{e}} 1}{(1)^{2}}$ $\Rightarrow \quad \mathrm{f}^{\prime}(\mathrm{e})=\frac{1}{2 \mathrm{e}}$

Hence, the correct answer is option: 'C'.

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