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If $y=\log _2\left(\log _2 x\right)$, then $\frac{d y}{d x}$ is equal to
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Verified Answer
The correct answer is:
$\frac{1}{\left(x \log _e x\right) \log _e 2}$
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
$\begin{aligned}
& \therefore \quad \frac{d y}{d x}=\frac{1}{\log 2}\left[\frac{1}{\log _e x} \times \frac{1}{x}-0\right]=\frac{1}{\log 2 \cdot \log _e x \cdot x} \\
& \frac{d y}{d x}=\frac{1}{\left(x \log _e x\right) \log _e 2}
\end{aligned}$
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
$\begin{aligned}
& \therefore \quad \frac{d y}{d x}=\frac{1}{\log 2}\left[\frac{1}{\log _e x} \times \frac{1}{x}-0\right]=\frac{1}{\log 2 \cdot \log _e x \cdot x} \\
& \frac{d y}{d x}=\frac{1}{\left(x \log _e x\right) \log _e 2}
\end{aligned}$
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