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?
 A: Your answer is correct, but you can simplify it further using the change of base formula. Where $\log x$ is in base $10$ and $\ln x$ is the natural log, you have
$$10^x\cdot \ln10\cdot \log x+\frac{1}{x\cdot \ln10}\cdot 10^x$$
$$=10^x\cdot \left(\ln10\cdot \color{green}{\log x}+\frac{1}{x\cdot \ln 10}\right)$$
$$=10^x\cdot \left(\ln10\cdot \color{green}{\frac{\ln x}{\ln 10}}+\frac{1}{x\cdot \ln 10}\right)$$
$$=10^x\cdot \left(\ln x+\frac{1}{x\cdot \ln 10}\right)$$
A: Hint. You may use
$$
(u \times v)'=u'v+uv'
$$ in the form
$$
\left(10^x\log_{10}(x)\right)'=\left(e^{x \ln 10} \times \frac{ \ln x}{\ln 10}\right)'.
$$ Can you take it from there?
A: Use the facts

$$(a^x)' = a^x \ln a, \quad (\log_a(x))' = \frac{1}{x\ln a}$$
  for $a > 0$ and $a \neq 1$.

Then use the multiplication formula for differentiation to get
\begin{align}
& (10^x \log_{10}x)' \\
= & (10^x)'\log_{10}x + 10^x (\log_{10}x)' \\
= & 10^x \ln (10) \log_{10}x + 10^x \frac{1}{x\ln 10}.
\end{align}
If you want to express your final result in natural logs, then you can simplify by writing 
$$\log_{10}x = \frac{\ln x}{\ln 10},$$
in this way the answer would agree with the wolfram output exactly.
A: Using 
$$\log_{10}(x) = \frac{\ln(x)}{\ln(10)}$$
then 
\begin{align}
D \left[ 10^{x} \, \log_{10}(x) \right] &= D \left[ e^{x \, \ln(10)} \, \frac{\ln(x)}{\ln(10)} \right] \\
&= \frac{1}{\ln(10)} \, \left[ \frac{10^{x}}{x} + \ln(10) \, 10^{x} \, \ln(x) \right] \\
&= 10^{x} \, \left[ \frac{1}{ x \, \ln(10)} + \ln(10) \, \log_{10}(x) \right]
\end{align}
A: \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.
A: $$\dfrac {\Bbb d} {\Bbb dx} [10x \dfrac {\ln x} {\ln 10}] =\dfrac {\frac {\Bbb d} {\Bbb dx} (10x \ln x)} {\ln 10} = \dfrac {\frac {\Bbb d} {\Bbb dx} [10^x] \ln x + 10^x \frac {\Bbb d} {\Bbb dx} \ln x} {\ln 10} = \dfrac {\ln 10 10^x \ln x + \frac 1 x 10^x} {\ln 10} = \dfrac {\ln 10 10^x \ln x + \frac {10^x} x} {\ln 10}$$
A: $$\frac{\text{d}}{\text{d}x}\left(10^x\cdot\log_{10}(x)\right)=\frac{\text{d}}{\text{d}x}\left(\frac{10^x\ln(x)}{\ln(10)}\right)=$$
$$\frac{\frac{\text{d}}{\text{d}x}\left(10^x\ln(x)\right)}{\ln(10)}=\frac{\ln(x)\frac{\text{d}}{\text{d}x}(10^x)+10^x\frac{\text{d}}{\text{d}x}(\ln(x))}{\ln(10)}=$$
$$\frac{10^x\ln(10)\ln(x)+10^x\frac{\text{d}}{\text{d}x}(\ln(x))}{\ln(10)}=\frac{10^x\ln(10)\ln(x)+\frac{10^x}{x}}{\ln(10)}=$$
$$10^x\left(\ln(x)+\frac{1}{x\ln(10)}\right)$$
