Question

Help!!differentiate0" id="TexFormula1" title=" { {e}^{x} }^{2} log_{10}(2x) " alt=" { {e}^{x} }^{2} log_{10}(2x) " align="absmiddle" class="latex-formula">
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Answer 1

Rewrite the function using the change-of-base identity as

$e^{x^2} \log_{10}(2x) = e^{x^2} \dfrac{\ln(2x)}{\ln(10)}$

Apply the product rule:

$\left(e^{x^2} \log_{10}(2x)\right)' = \left(e^{x^2}\right)' \dfrac{\ln(2x)}{\ln(10)} + e^{x^2} \left(\dfrac{\ln(2x)}{\ln(10)}\right)'$

Use the chain rule:

$\left(e^{x^2} \log_{10}(2x)\right)' = e^{x^2}\left(x^2\right)' \dfrac{\ln(2x)}{\ln(10)} + e^{x^2} \dfrac{(2x)'}{2\ln(10)x}$

Compute the remaining derivatives:

$\left(e^{x^2} \log_{10}(2x)\right)' = 2xe^{x^2} \dfrac{\ln(2x)}{\ln(10)} + e^{x^2} \dfrac2{2\ln(10)x} = e^{x^2}\left(\dfrac{2x\ln(2x)}{\ln(10)} + \dfrac1{\ln(10)x}\right)$

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

$\left(e^{x^2} \log_{10}(2x)\right)' = e^{x^2}\left(2x\log_{10}(2x)+\frac{\log_{10}(e)}x\right)$