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

Hint: You'll need to use the product rule with $2x$ and the logarithmic function.
Then, whilst applying the product rule, you use the chain rule on the logarithmic function. 
Notice: differentiating $\log_a f(x)$ gives you (this is where you made your mistake) $$(\log_a f(x))' = \frac{f'(x)}{\ln a \cdot f(x)}$$
Use that with $f(x) = \sqrt{x}$. In fact, you should get $$\bbox[border: solid blue 1px, 10px]{(\log_{10} \sqrt{x})' = \frac{1}{2x \ln 10}}$$

Full solution:
Hence, using the product rule we have the derivative as $$2x \cdot \frac{1}{2x \ln 10} + 2 \cdot \frac{1}{2x \ln 10} = \frac{1}{\ln 10} + \frac{1}{x \ln 10}$$
