Q269 / 270 · NDA 2016 (Phase 1) · Mathematics
If $\mathrm{y}=\log _{10} \mathrm{x}+\log _{\mathrm{x}} 10+\log _{\mathrm{x}} \mathrm{x}+\log _{10} 10$ then what is $\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{\mathrm{x}=10}$ equal to?
- A10
- B2
- C1
- Correct answerD0
Solution
$\begin{aligned}
& y=\log _{10} x+\log _x 10+\log _x x+\log _{10} 10 \\
& y=\log _{10} x+\log _x 10+1+1
\end{aligned}$
Differentiating equation w.r.t. x
$\begin{aligned}
& \frac{d y}{d x}=\frac{1}{x \log _e 10}-\frac{1}{\left(\log _{10} x\right)^2} \cdot \frac{1}{(x \log 10)} \\
& =\frac{1}{x \log _e 10}\left[1-\frac{1}{\left(\log _{10} x\right)^2}\right] \\
& \left(\frac{d y}{d x}\right)_{x=10}=\frac{1}{10 \log _e 10}[1-1]=0 \\
& {\left[\begin{array}{rl}
\text { Note: } \log _x 10 & =\frac{\log _{10} 10}{\log _{10} x}=\frac{1}{\log _{10} x} \\
\frac{d}{d x}\left[\frac{1}{\log _{10} x}\right] & =-\left(\log _{10} x\right)^{-2} \times \frac{1}{x \log _e 10} \\
& =-\frac{1}{\left(\log _{10} x\right)^2 x \log _e 10}
\end{array}\right]}
\end{aligned}$
Hence, the correct answer is option: 'D'.