Question

alguém ai consegue derivar???
são funções
$y = {x}^{ {x}^{x} }$
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Answer 1

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$\sf y=x^{x^x}\\\sf \ell n y=\ell nx^{x^x}\\\sf \ell ny=x^x\ell nx\\\sf c\acute alculo~da~derivada~de~x^x:\\\sf y=x^x=e^{x\ell nx}\\\sf \dfrac{dy}{dx}= e^{x\ell nx}\cdot(x\cdot\ell nx)'\\\sf \dfrac{dy}{dx}=e^{x\ell nx}\cdot\left(\ell nx+\diagup\!\!\!x\cdot\dfrac{1}{\diagup\!\!\!x}\right)\\\sf \dfrac{dy}{dx}=e^{x\ell nx}\cdot (\ell nx+1)\\\sf \dfrac{dy}{dx}=x^x\cdot(\ell nx+1)$

$\sf\ell ny=x^x\cdot\ell nx\\\sf\dfrac{dy}{dx}\cdot\dfrac{1}{y}=(x^x)'\cdot\ell nx+x^x\cdot (\ell nx)'\\\sf substituindo~a~derivada~de~x^x~na~express\tilde ao~temos:\\\sf \dfrac{dy}{dx}\cdot\dfrac{1}{y}=[x^x\cdot(\ell nx+1)]\cdot\ell nx+x^x\cdot\dfrac{1}{x}\\\sf \dfrac{dy}{dx}\cdot\dfrac{1}{y}=x^x\ell nx(\ell nx+1)+x^{x-1}\\\sf isolando\dfrac{dy}{dx}~temos:$

$\sf \dfrac{dy}{dx}=y[x^x\ell nx(\ell nx+1)+x^{x-1}]\\\sf substituindo~y~na~express\tilde ao~temos:\\\large\boxed{\boxed{\boxed{\boxed{\sf\dfrac{dy}{dx}=x^{x^x}\cdot[x^x\ell nx(\ell nx+1)+x^{x-1}]}}}}$