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

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(A) Given,$y = \log_{10} x + \log_{x} 10 + \log_{x} x + \log_{10} 10$. Using the change of base formula $\log_{a} b = \frac{\log_{e} b}{\log_{e} a}$,w

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) 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 can rewrite the expression as: $y = \frac{\log_{e} x}{\log_{e} 10} + \frac{\log_{e} 10}{\log_{e} x} + 1 + 1$. $y = \frac{1}{\log_{e} 10} \cdot \log_{e} x + \log_{e} 10 \cdot (\log_{e} x)^{-1} + 2$. Now,differentiate 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 $\frac{dy}{dx}$ is equal to

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