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 $y'= \frac{1}{\ln10}+\log_{10}x$

Firstly, note that $$\log_{10} \sqrt{x} = \dfrac{1}{2} \log_{10} x$$
Then we have that $$ \begin{aligned} f(x) = \log_{10} (x) & \iff 10^{f(x)} = x \\ & \implies f(x) \log 10 = \log x \\ & \implies \dfrac{\text{d}}{\text{d}x} f(x) \log 10 = \dfrac{\text{d}}{\text{d}x} \\ & \implies f'(x) \log 10 = \dfrac{1}{x} \\ & \implies f'(x) = \dfrac{1}{x \log 10} \end{aligned} $$
Use the product rule and you should get your answer.
(Note that the above method for finding the derivative applies to any arbitrary base $0 < a \in \mathbb{R}$)
