If $y=\log _{10} x+\log _x 10+\log _x x+\log _{10} 10$, then $\frac{d y}{d x}=$

If $y=\log _{10} x+\log _x 10+\log _x x+\log _{10} 10$, then $\frac{d y}{d x}=$
  1. $\frac{1}{x \log _e}+\frac{1}{x \log _{10} e}$
  2. $\frac{1}{x \log _e 10}+\frac{\log _e 10}{x\left(\log _{10} e\right)^2}$
  3. $\frac{1}{x \log _e10}+\frac{1}{x \log _{10} e}$
  4. $\frac{1}{x \log _e 10}-\frac{\log _0 10}{x\left(\log _e x\right)^2}$

Solution

$\begin{aligned} & y=\frac{\log _e x}{\log _e 10}+\frac{\log _e 10}{\log _e x}+1+1 \\ & \therefore \frac{d y}{d x}=\left(\frac{1}{\log _e 10}\right)\left(\frac{1}{x}\right)+\left(\log _e 10\right)\left[\frac{-\frac{1}{x}}{\left(\log _e x\right)^2}\right] \\ & =\frac{1}{x \log _e 10}-\frac{\log _e 10}{x\left(\log _e x\right)^2}\end{aligned}$

Asked in: MHT CET 2021 (23 Sep Shift 1)

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