Question

Derivada de
$y=ln( \frac{4}{x}+1)$

Answer 1

$y=\mathrm{\ell n}\left(\dfrac{4}{x}+1\right)$


Derivaremos usando a Regra da Cadeia, combinada com as outras regras de derivação:

$\dfrac{dy}{dx}=\dfrac{d}{dx}\!\left[\mathrm{\ell n}\left(\dfrac{4}{x}+1\right) \right ]\\\\\\ =\dfrac{1}{\frac{4}{x}+1}\cdot \dfrac{d}{dx}\!\left(\dfrac{4}{x}+1 \right )\\\\\\ =\dfrac{1}{\frac{4}{x}+1}\cdot \dfrac{d}{dx}\,(4x^{-1}+1)\\\\\\ =\dfrac{1}{\frac{4}{x}+1}\cdot (-4x^{-2}+0)\\\\\\ =\dfrac{1}{\frac{4}{x}+1}\cdot \left(-\,\dfrac{4}{x^{2}}\right)$

$=-\,\dfrac{4}{x^2 \cdot \left(\frac{4}{x}+1\right)}\\\\\\ \therefore~~\boxed{\begin{array}{c} \dfrac{dy}{dx}=-\,\dfrac{4}{4x+x^2} \end{array}}$

Answer 2

Temos a seguinte derivada:

$y = Ln( \frac{4}{x} +1)$

Deveremos aplicar a regra da cadeia. onde:

$Y' = [ Ln( \frac{4}{x} +1 )]'*( \frac{4}{x} +1)'$

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Observe que derivada de Ln é 1/logaritmando 

Então:



$\\ Y' = \frac{1}{ \frac{4}{x}+1 } *(\frac{4}{x}+1 } )' \\ \\ Y' = \frac{1}{ \frac{4}{x} +1} *(4*x^-^1+1)' \\ \\ Y' = \frac{1}{ \frac{4}{x} +1}*(-1*4*x^-^1^-^1+0) \\ \\ Y' = \frac{1}{ \frac{4}{x}+1 }*(-4x^-^2) \\ \\ Y' = \frac{-4x^-^2}{ \frac{4}{x} +1}$

Passe invertido a fração de baixo: :)


$\\ Y' = -4x^-^2* \frac{x}{x+4} \\ \\ Y' = -1*x^-^2^+^1/(x+4) \\ \\ Y' = -x^-^1/(x+4) \\ \\ Y' = - \frac{1}{x^2+4x}$