Rewrite the function using the change-of-base identity as
[tex]e^{x^2} \log_{10}(2x) = e^{x^2} \dfrac{\ln(2x)}{\ln(10)}[/tex]
Apply the product rule:
[tex]\left(e^{x^2} \log_{10}(2x)\right)' = \left(e^{x^2}\right)' \dfrac{\ln(2x)}{\ln(10)} + e^{x^2} \left(\dfrac{\ln(2x)}{\ln(10)}\right)'[/tex]
Use the chain rule:
[tex]\left(e^{x^2} \log_{10}(2x)\right)' = e^{x^2}\left(x^2\right)' \dfrac{\ln(2x)}{\ln(10)} + e^{x^2} \dfrac{(2x)'}{2\ln(10)x}[/tex]
Compute the remaining derivatives:
[tex]\left(e^{x^2} \log_{10}(2x)\right)' = 2xe^{x^2} \dfrac{\ln(2x)}{\ln(10)} + e^{x^2} \dfrac2{2\ln(10)x} = e^{x^2}\left(\dfrac{2x\ln(2x)}{\ln(10)} + \dfrac1{\ln(10)x}\right)[/tex]
If you like, you can convert back to base-10 logarithms:
ln(2x) / ln(10) = log₁₀(2x)
1 / ln(10) = ln(e) / ln(10) = log₁₀(e)
Then
[tex]\left(e^{x^2} \log_{10}(2x)\right)' = e^{x^2}\left(2x\log_{10}(2x)+\frac{\log_{10}(e)}x\right)[/tex]
