If y = log_{10}x + log_x 10 + log_x x + log_{ | vedclass

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(A) Given the function: $y = \log_{10}x + \log_x 10 + \log_x x + \log_{10} 10$ Using the change of base formula $\log_a b = \frac{\ln b}{\ln a}$,we si

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Given the function: $y = \log_{10}x + \log_x 10 + \log_x x + \log_{10} 10$ Using the change of base formula $\log_a b = \frac{\ln b}{\ln a}$,we simplify the expression: $y = \frac{\ln x}{\ln 10} + \frac{\ln 10}{\ln x} + 1 + 1$ Now,differentiate with respect to $x$: $\frac{dy}{dx} = \frac{d}{dx} \left( \frac{\ln x}{\ln 10} \right) + \frac{d}{dx} \left( \frac{\ln 10}{\ln x} \right) + \frac{d}{dx}(2)$ $\frac{dy}{dx} = \frac{1}{\ln 10} \cdot \frac{1}{x} + \ln 10 \cdot \frac{d}{dx} ((\ln x)^{-1}) + 0$ $\frac{dy}{dx} = \frac{1}{x \ln 10} + \ln 10 \cdot (-1)(\ln x)^{-2} \cdot \frac{1}{x}$ $\frac{dy}{dx} = \frac{1}{x \ln 10} - \frac{\ln 10}{x (\ln x)^2}$ Thus,the correct option is $A$.

If $y = \log_{10}x + \log_x 10 + \log_x x + \log_{10} 10$,then $\frac{dy}{dx} = $

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