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If $y=\log _{a} x+\log _{x} a+\log _{x} x+\log _{a} a$, then $\frac{d y}{d x}$ is equal to
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Verified Answer
The correct answer is:
$\frac{1}{x \log a}-\frac{\log a}{x(\log x)^{2}}$
We have,
$\begin{aligned}
y &=\log _{a} x+\frac{\log a}{\log x}+1+1 \quad\left[\because \log _{x} x=1\right] \\
\Rightarrow \frac{d y}{d x} &=\frac{1}{x} \log _{a} e-\log a\left(\frac{1}{\log x}\right)^{2} \frac{1}{x}=\frac{1}{x \log a}-\frac{\log a}{x(\log x)^{2}}
\end{aligned}$
$\begin{aligned}
y &=\log _{a} x+\frac{\log a}{\log x}+1+1 \quad\left[\because \log _{x} x=1\right] \\
\Rightarrow \frac{d y}{d x} &=\frac{1}{x} \log _{a} e-\log a\left(\frac{1}{\log x}\right)^{2} \frac{1}{x}=\frac{1}{x \log a}-\frac{\log a}{x(\log x)^{2}}
\end{aligned}$
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