frac{d}{dx}(e^{x log x} + e^3) = . . . . . | vedclass

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(A) Let $y = e^{x \log x} + e^3$. Since $e^{x \log x} = e^{\log(x^x)} = x^x$,the expression becomes $y = x^x + e^3$. Now,differentiate with respect to

Vedclass · Accepted Answer

$x^x(1 + \log x)$

Vedclass · Answer

(A) Let $y = e^{x \log x} + e^3$. Since $e^{x \log x} = e^{\log(x^x)} = x^x$,the expression becomes $y = x^x + e^3$. Now,differentiate with respect to $x$: $\frac{dy}{dx} = \frac{d}{dx}(x^x) + \frac{d}{dx}(e^3)$. Since $e^3$ is a constant,its derivative is $0$. To differentiate $x^x$,let $u = x^x$. Then $\log u = x \log x$. Differentiating both sides with respect to $x$: $\frac{1}{u} \frac{du}{dx} = 1 \cdot \log x + x \cdot \frac{1}{x} = \log x + 1$. Thus,$\frac{du}{dx} = u(1 + \log x) = x^x(1 + \log x)$. Therefore,$\frac{dy}{dx} = x^x(1 + \log x) + 0 = x^x(1 + \log x)$.

$\frac{d}{dx}(e^{x \log x} + e^3) = $ . . . . . .

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