Derivada implitita

por **Diego Silva** » Dom Jul 28, 2013 13:12

Pessoal gostaria de ajuda para encontrar a segunda derivada de cos y = x e 3x² + 4y² = 4:

A primeira consigo encontrar mas a segunda não consigo, se alguém puder ajudar.

por **MateusL** » Dom Jul 28, 2013 16:41

A segunda ficará:
$\dfrac{d(3x^2+4y^2)}{dx}=0$

$3\cdot\dfrac{dx^2}{dx}+4\cdot\dfrac{dy^2}{dx}=0$

$6x+8y\dfrac{dy}{dx}=0$

$\dfrac{dy}{dx}=-\dfrac{3x}{4y}$

$\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}\left(-\dfrac{3x}{4y}\right)=-\dfrac{3}{4}\left(x\dfrac{d}{dx}\left(\dfrac{1}{y}\right)+\dfrac{1}{y}\cdot \dfrac{dx}{dx}\right)=-\dfrac{3}{4}\left(x\cdot\dfrac{-\frac{dy}{dx}}{y^2}+\dfrac{1}{y}\right)$

$\dfrac{d^2y}{dx^2}=\dfrac{3}{4}\left(\dfrac{x}{y^2}\cdot \dfrac{dy}{dx}-\dfrac{1}{y}\right)=\dfrac{3}{4}\left(\dfrac{x}{y^2}\cdot\dfrac{-3x}{4y}-\dfrac{1}{y}\right)$

$\dfrac{d^2y}{dx^2}=-\dfrac{3}{4}\left(\dfrac{3x^2}{4y^3}+\dfrac{1}{y}\right)=-\dfrac{3}{4}\left(\dfrac{3x^2+4y^2}{4y^3}\right)=-\dfrac{3}{4}\left(\dfrac{4}{4y^3}\right)$

$\dfrac{d^2y}{dx^2}=-\dfrac{3}{4y^3}$

Acredito que seja isso.

Abraço!

por **Diego Silva** » Seg Jul 29, 2013 20:54

MateusL escreveu:A segunda ficará:
$\dfrac{d(3x^2+4y^2)}{dx}=0$

$3\cdot\dfrac{dx^2}{dx}+4\cdot\dfrac{dy^2}{dx}=0$

$6x+8y\dfrac{dy}{dx}=0$

$\dfrac{dy}{dx}=-\dfrac{3x}{4y}$

$\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}\left(-\dfrac{3x}{4y}\right)=-\dfrac{3}{4}\left(x\dfrac{d}{dx}\left(\dfrac{1}{y}\right)+\dfrac{1}{y}\cdot \dfrac{dx}{dx}\right)=-\dfrac{3}{4}\left(x\cdot\dfrac{-\frac{dy}{dx}}{y^2}+\dfrac{1}{y}\right)$

$\dfrac{d^2y}{dx^2}=\dfrac{3}{4}\left(\dfrac{x}{y^2}\cdot \dfrac{dy}{dx}-\dfrac{1}{y}\right)=\dfrac{3}{4}\left(\dfrac{x}{y^2}\cdot\dfrac{-3x}{4y}-\dfrac{1}{y}\right)$

$\dfrac{d^2y}{dx^2}=-\dfrac{3}{4}\left(\dfrac{3x^2}{4y^3}+\dfrac{1}{y}\right)=-\dfrac{3}{4}\left(\dfrac{3x^2+4y^2}{4y^3}\right)=-\dfrac{3}{4}\left(\dfrac{4}{4y^3}\right)$

$\dfrac{d^2y}{dx^2}=-\dfrac{3}{4y^3}$

Acredito que seja isso.

Abraço!

isso mesmo, obrigado!

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