If $y = 3^x + e^x + x^x + x^3$, then the value of $\frac{dy}{dx}$ at $x = 3$ is

Given $y = 3^x + e^x + x^x + x^3$

Finding $\frac{dy}{dx}$ by differentiating each term:

$\frac{d}{dx}(3^x) = 3^x \log_e 3$

$\frac{d}{dx}(e^x) = e^x$

$\frac{d}{dx}(x^x) = x^x(\log_e x + 1)$

$\frac{d}{dx}(x^3) = 3x^2$

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Combining all derivatives:

$\frac{dy}{dx} = 3^x \log_e 3 + e^x + x^x(\log_e x + 1) + 3x^2$

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Substituting $x = 3$:

$\frac{dy}{dx}\bigg|_{x=3} = 3^3 \log_e 3 + e^3 + 3^3(\log_e 3 + 1) + 3(3)^2$

$= 27\log_e 3 + e^3 + 27(\log_e 3 + 1) + 27$

$= 27\log_e 3 + e^3 + 27\log_e 3 + 27 + 27$

$= e^3 + 54\log_e 3 + 54$

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Therefore, the value of $\frac{dy}{dx}$ at $x = 3$ is:

$e^3 + 54\log_e 3 + 54$