Question

Como derivar uma função como essa y=ln(x+raiz de X ao quadrado -1)

Answer 1

Utilizando a regra da cadeia:

$\frac{dy}{dx}=\frac{dy}{du}.\frac{du}{dx}$

Sejam

$y=\ln(u)\\\\u=x+\sqrt{x^2}-1$

$\frac{d}{dx}\ln(x+\sqrt{x^2}-1)=\frac{d}{du}\ln(u).\frac{d}{dx}(x+\sqrt{x^2}-1)=\frac{d}{du}\ln(u).\frac{d}{dx}(x+|x|-1)\\\\(\frac{1}{u}).(1+\frac{x}{|x|})\\\\\frac{1}{u}+\frac{x}{u.|x|}\\\\\frac{1}{x+|x|-1}+\frac{x}{(1+|x|-1)|x|}\\\\\frac{|x|+x}{(1+|x|-1)|x|}$

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$\frac{d}{dx}(c)=0\\\\\frac{d}{dx}(x)=1\\\\\frac{d}{dx}(\ln(x))=\frac{1}{x}\\\\\frac{d}{dx}(|x|)=\frac{x}{|x|}$