$x$ এর সাপেক্ষে অন্তরক সহগ নিচের কোনটি? $\log _{a} x+\log _{x} a$

Solve:

$\begin{aligned} & \log _{a} x+\log _{x} a \\ = & \log _{a} e \times \log _{e} x+\log _{x} e \times \log _{e} a \\ = & \frac{1}{\log _{e} a} \times \ln x+\frac{1}{\log _{e} x} \times \ln a \\ = & \frac{1}{\ln a} \times \ln x+\ln a \times(\ln x)^{-1} \\ \therefore & \frac{d}{d x}\left(\log _{a} x+\log _{x} a\right) \\ = & \frac{1}{\ln a} \frac{1}{x}+\ln a \times\left\{-1(\ln x)^{-2} \frac{1}{x}\right\} \\ = & \frac{1}{x \ln a}-\frac{\ln a}{x(\ln x)^{2}} \end{aligned}$