<s>[INST] Derivative of $10^x\cdot\log_{10}(x)$ 
Derive  $10^x\cdot\log_{10}(x)$ 

$$10^x\cdot \ln(10)\cdot \log_{10}(x)+\frac{1}{x\cdot \ln(10)}\cdot 10^x$$
But WolframAlpha gives another solution. Where am I wrong?
 [/INST] \begin{align}
\text{your answer} & = 10^x\cdot \ln10\cdot \log_{10} x+\frac{1}{x\cdot \ln10}\cdot 10^x \\[10pt]
& = 10^x \ln x + \frac{10^x}{x\ln 10} \\[10pt]
& = 10^x \left( \ln x + \frac 1 {x\ln 10} \right) = \text{Wolfram's answer}.
\end{align}
Note that where Wolfram writes $\log x$ or $\log 10$ with no base specified, it means the base is $e$, so it's the natural logarithm.
Note also that we used the identity $\ln10\cdot\log_{10}x = \ln x$.  That is an instance of the change-of-base formula for logarithms.
</s>