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 $y'= \frac{1}{\ln10}+\log_{10}x$
 A: Your derivative of $\log_{10} \sqrt{x}$ is incorrect. I recommend having another go and posting your working here for correction if you get the same thing.
A: 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 $$2x \cdot \frac{1}{2x \ln 10} + 2 \cdot \frac{1}{2x \ln 10} = \frac{1}{\ln 10} + \frac{1}{x \ln 10}$$
A: $$\log\sqrt{x} = \frac{1}{2}\log_{10} x= \frac{\ln x}{2\ln 10}$$
$$\frac{d}{dx}\ln x= \frac{1}{x}$$
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: $$\log_{10}x = \frac{\ln x}{\ln 10} $$
A: Notice, $$\frac{d}{dx}(\log_{a}(x))=\frac{d}{dx}(\log_{a}(e)(\log_{e}(x)))$$ Now, we have $$y=2x\log_{10}\sqrt{x}$$ $$\implies y=x\log_{10}(\sqrt{x})^2$$ $$\implies y=x\log_{10}(x)$$  $$\implies y=x\log_{e}(x)\times \log_{10}(e)$$ $$\implies \frac{dy}{dx}=\log_{10}(e)\frac{d}{dx}\left(x\log_{e}(x)\right)$$ $$\implies \frac{dy}{dx}=\log_{10}(e)\left(x\frac{1}{x}+(1)\log_{e}(x)\right)$$  $$\implies \frac{dy}{dx}=\log_{10}(e)+\log_{10}(e)\times \log_{e}(x)$$  $$\implies \frac{dy}{dx}=\log_{10}(e)+\log_{10}(x)$$  $$\implies \frac{dy}{dx}=\log_{10}(ex)$$ 
A: 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 $0 < a \in \mathbb{R}$)
A: Take $\log_{10}x$ as $\log x$.
$$y=2x\log\sqrt{x}=x\log x=\log x^x$$
$$10^y=x^x \implies y\ln10=x\ln x \implies y=\frac{x\ln x}{ \ln10}$$
Then
$$y'=\frac{1}{\ln10}\times(\ln x + 1)=\log x + \frac{1}{\ln 10}.$$
A: Here are the steps
$$\frac{d}{dx}\left[2x\log_{10}\sqrt{x}\right]$$
$$=\frac{d}{dx}\left[2x\log_{10} x^{\frac12}\right]$$
$$=\frac{d}{dx}\left[\frac22x\log_{10} x\right]$$
$$=\frac{d}{dx}\left[x\log_{10} x\right]$$
$$=\frac{d}{dx}\left[\frac{x\log x}{\log 10}\right]$$
$$=\frac{1}{\log 10}\frac{d}{dx}\left[x\log x\right]$$
$$=\frac{1}{\log 10}\left(x\frac{d}{dx}\left[\log x\right]+(\log x)\frac{d}{dx}[x]\right)$$
$$=\frac{1}{\log 10}\left(\frac{x}{x}+\log x\right)$$
$$=\frac{1+\log x}{\log 10}$$
Which is equivalent to the books answer.
