<s>[INST] Differentiate the Function: $y=2x \log_{10}\sqrt{x}$ $y=2x\log_{10}\sqrt{x}$
Solve using: Product Rule $\left(f(x)\cdot g(x)\right)'= f(x)\cdot\frac{d}{dx}g(x)+g(x)\cdot \frac{d}{dx}f(x)$
and $\frac{d}{dx}(\log_ax)= \frac{1}{x\ \ln\ a}$
$(2x)\cdot [\log_{10}\sqrt{x}]'+(\log_{10}\sqrt{x})\cdot [2x]'$
$y'=2x\frac{1}{\sqrt{x}\ln 10}+\log_{10}\sqrt{x}\cdot 2$
Answer in book is $y'= \frac{1}{\ln10}+\log_{10}x$
 [/INST] $$\log\sqrt{x} = \frac{1}{2}\log_{10} x= \frac{\ln x}{2\ln 10}$$
$$\frac{d}{dx}\ln x= \frac{1}{x}$$
using chain rule,
$$\frac{d}{dx}\left(2x\log\sqrt{x}\right)= 2x \frac{dx}{dx}\frac{\ln x}{2\ln 10} + \frac{\ln x}{2\ln 10}\frac{d}{dx} 2x $$
$$=\frac{2x}{2x\ln 10} + \frac{2\ln x}{2\ln 10} $$
$$ =\frac{1}{\ln 10}+ \log_{10} x$$
Note the change of base: $$\log_{10}x = \frac{\ln x}{\ln 10} $$
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