left{frac{d}{d x}left(x^x+x^{x+1}+x^{x+2}righ | vedclass

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Question

(C) Let $f(x) = x^x + x^{x+1} + x^{x+2}$.We know that $\frac{d}{dx}(x^x) = x^x(1 + \ln x)$.For $x^{x+1}$,let $y = x^{x+1}$,then $\ln y = (x+1)\ln x$.

Vedclass · Accepted Answer

$e^e(2e^2+4e+3)$

Vedclass · Answer

(C) Let $f(x) = x^x + x^{x+1} + x^{x+2}$.We know that $\frac{d}{dx}(x^x) = x^x(1 + \ln x)$.For $x^{x+1}$,let $y = x^{x+1}$,then $\ln y = (x+1)\ln x$. Differentiating both sides: $\frac{1}{y} \frac{dy}{dx} = \ln x + \frac{x+1}{x} = \ln x + 1 + \frac{1}{x}$. So,$\frac{d}{dx}(x^{x+1}) = x^{x+1}(\ln x + 1 + \frac{1}{x})$.For $x^{x+2}$,let $y = x^{x+2}$,then $\ln y = (x+2)\ln x$. Differentiating both sides: $\frac{1}{y} \frac{dy}{dx} = \ln x + \frac{x+2}{x} = \ln x + 1 + \frac{2}{x}$. So,$\frac{d}{dx}(x^{x+2}) = x^{x+2}(\ln x + 1 + \frac{2}{x})$.Now,evaluate at $x=e$:$\frac{d}{dx}(x^x)|_{x=e} = e^e(1 + \ln e) = e^e(1+1) = 2e^e$.$\frac{d}{dx}(x^{x+1})|_{x=e} = e^{e+1}(\ln e + 1 + \frac{1}{e}) = e^{e+1}(2 + \frac{1}{e}) = 2e^{e+1} + e^e = e^e(2e+1)$.$\frac{d}{dx}(x^{x+2})|_{x=e} = e^{e+2}(\ln e + 1 + \frac{2}{e}) = e^{e+2}(2 + \frac{2}{e}) = 2e^{e+2} + 2e^{e+1} = e^e(2e^2+2e)$.Summing these: $e^e(2 + 2e + 1 + 2e^2 + 2e) = e^e(2e^2 + 4e + 3)$.

$\left\{\frac{d}{d x}\left(x^x+x^{x+1}+x^{x+2}\right)\right\}_{x=e} = \text{?}$

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