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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?
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$\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{\mathrm{x}=10}$ equal to?
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$y=\log _{10} x+\log _{x} 10+\log _{x} x+\log _{10} 10$
$y=\log _{10} x+\log _{x} 10+1+1$
Differentiating equation w.r.t. $\mathrm{x}$
$\frac{d y}{d x}=\frac{1}{x \log _{e} 10}-\frac{1}{\left(\log _{10} x\right)^{2}} \cdot \frac{1}{(x \log 10)}$
$=\frac{1}{x \log _{e} 10}\left[1-\frac{1}{\left(\log _{10} x\right)^{2}}\right]$
$\left(\frac{d y}{d x}\right)_{x=10}=\frac{1}{10 \log _{e} 10}[1-1]=0$
$\left[\begin{array}{l}\text { Note: } \log _{x} 10=\frac{\log _{10} 10}{\log _{10} x}=\frac{1}{\log _{10} x} \\ \frac{d}{d x}\left[\frac{1}{\log _{10} x}\right]=-\left(\log _{10} x\right)^{-2} \times \frac{1}{x \log _{e} 10} \\ =-\frac{1}{\left(\log _{10} x\right)^{2} x \log _{e} 10}\end{array}\right]$
$y=\log _{10} x+\log _{x} 10+1+1$
Differentiating equation w.r.t. $\mathrm{x}$
$\frac{d y}{d x}=\frac{1}{x \log _{e} 10}-\frac{1}{\left(\log _{10} x\right)^{2}} \cdot \frac{1}{(x \log 10)}$
$=\frac{1}{x \log _{e} 10}\left[1-\frac{1}{\left(\log _{10} x\right)^{2}}\right]$
$\left(\frac{d y}{d x}\right)_{x=10}=\frac{1}{10 \log _{e} 10}[1-1]=0$
$\left[\begin{array}{l}\text { Note: } \log _{x} 10=\frac{\log _{10} 10}{\log _{10} x}=\frac{1}{\log _{10} x} \\ \frac{d}{d x}\left[\frac{1}{\log _{10} x}\right]=-\left(\log _{10} x\right)^{-2} \times \frac{1}{x \log _{e} 10} \\ =-\frac{1}{\left(\log _{10} x\right)^{2} x \log _{e} 10}\end{array}\right]$
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