Q107 / 160 · AP EAMCET 2016 · Mathematics

If $y=\log _2\left(\log _2 x\right)$, then $\frac{d y}{d x}$ is equal to

  • A
    $\frac{\log _e 2}{x \log _e x}$
  • B
    $\frac{1}{\log _e(2 x)^x}$
  • Correct answer
    C
    $\frac{1}{\left(x \log _e x\right) \log _e 2}$
  • D
    $\frac{1}{x\left(\log _2 x\right)^2}$

Solution

We have, $y=\log _2\left(\log _2 x\right)$ Since, $\log _a^b=\frac{\log b}{\log a}$ $\begin{aligned} & \therefore \quad y=\log _2\left(\frac{\log x}{\log 2}\right) \\ & \therefore \quad y=\frac{\log \frac{\log x}{\log 2}}{\log 2} \\ & \Rightarrow \quad y=\frac{\log (\log x)-\log (\log 2)}{\log 2} \end{aligned}$ On differentiating w.r.f $x$, we get $\begin{aligned} & \therefore \quad \frac{d y}{d x}=\frac{1}{\log 2}\left[\frac{1}{\log _e x} \times \frac{1}{x}-0\right]=\frac{1}{\log 2 \cdot \log _e x \cdot x} \\ & \frac{d y}{d x}=\frac{1}{\left(x \log _e x\right) \log _e 2} \end{aligned}$

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

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