If y=log _{10} x+log _x 10+log _x x+log _{10} | vedclass

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(B) Given $y = \log_{10} x + \log_x 10 + \log_x x + \log_{10} 10$. Using the change of base formula $\log_a b = \frac{\log_e b}{\log_e a}$,we have: $y

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$\frac{1}{x \log _e 10}-\frac{\log _e 10}{x(\log _e x)^2}$

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(B) Given $y = \log_{10} x + \log_x 10 + \log_x x + \log_{10} 10$. Using the change of base formula $\log_a b = \frac{\log_e b}{\log_e a}$,we have: $y = \frac{\log_e x}{\log_e 10} + \frac{\log_e 10}{\log_e x} + 1 + 1$. $y = \frac{\log_e x}{\log_e 10} + \frac{\log_e 10}{\log_e x} + 2$. Differentiating with respect to $x$: $\frac{dy}{dx} = \frac{1}{\log_e 10} \cdot \frac{d}{dx}(\log_e x) + \log_e 10 \cdot \frac{d}{dx}((\log_e x)^{-1}) + 0$. $\frac{dy}{dx} = \frac{1}{\log_e 10} \cdot \frac{1}{x} + \log_e 10 \cdot (-1)(\log_e x)^{-2} \cdot \frac{1}{x}$. $\frac{dy}{dx} = \frac{1}{x \log_e 10} - \frac{\log_e 10}{x(\log_e x)^2}$.

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

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