Question

Me ajudem a responder esse trabalho por favor!

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Answer 1

Em ambos os casos será utilizada a derivação implícita
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a)

É interessante acharmos a derivada de y em relação a x, que nos dá a inclinação da reta tangente em qualquer ponto.

Considerando que a equação da elipse pode ser escrita como uma função perto do ponto em questão, vamos derivar sua equação implicitamente em relação a x:

$\dfrac{2\cdot x^{2-1}}{9}\dfrac{dx}{dx}+\dfrac{2\cdot y^{2-1}}{4}\dfrac{dy}{dx}=0\\\\\\\dfrac{2x}{9}+\dfrac{y}{2}\dfrac{dy}{dx}=0\\\\\\\dfrac{y}{2}\cdot\dfrac{dy}{dx}=-\dfrac{2x}{9}\\\\\\\dfrac{dy}{dx}=-\dfrac{2x\cdot2}{9\cdot y}\\\\\\\boxed{\boxed{y'=-\dfrac{4x}{9y}}}$

Vamos achar a derivada no ponto dado (inclinação da reta tangente à elipse no ponto):

$y'=\dfrac{-4x}{9y}\\\\\\y'=\dfrac{-4(-1)}{9(-\frac{4\sqrt{2}}{3})}\\\\\\y'=\dfrac{4}{3(-4\sqrt{2})}\\\\\\y'=\dfrac{1}{-3\sqrt{2}}\\\\\\y'=-\dfrac{\sqrt{2}}{3\sqrt{2}\sqrt{2}}\\\\\\\boxed{\boxed{y'=-\dfrac{\sqrt{2}}{6}}}$

Achando a equação da reta tangente ao gráfico nesse ponto:

$y-y_{p}=y'(x-x_{p})\\\\\\y-\left(-\dfrac{4\sqrt{2}}{3}\right)=\left(-\dfrac{\sqrt{2}}{6}\right)(x-[-1])\\\\\\y+\dfrac{4\sqrt{2}}{3}=-\dfrac{\sqrt{2}}{6}x-\dfrac{\sqrt{2}}{6}\\\\\\y=-\dfrac{\sqrt{2}}{6}x-\dfrac{\sqrt{2}}{6}-\dfrac{4\sqrt{2}}{3}\\\\\\y=-\dfrac{\sqrt{2}}{6}x-\dfrac{\sqrt{2}}{6}-\dfrac{8\sqrt{2}}{6}\\\\\\y=-\dfrac{\sqrt{2}}{6}x+\dfrac{-\sqrt{2}-8\sqrt{2}}{6}\\\\\\y=-\dfrac{\sqrt{2}}{6}x+\dfrac{(-9\sqrt{2})}{6}\\\\\\\boxed{\boxed{y=-\dfrac{\sqrt{2}}{6}x-\dfrac{9\sqrt{2}}{6}}}$
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2) a)

$x^{\frac{1}{2}}+y^{\frac{1}{2}}=1\\\\\\\dfrac{1}{2}x^{\frac{1}{2}-1}\dfrac{dx}{dx}+\dfrac{1}{2}y^{\frac{1}{2}-1}\dfrac{dy}{dx}=0\\\\\\x^{-\frac{1}{2}}+y^{-\frac{1}{2}}\dfrac{dy}{dx}=0\\\\\\\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}\dfrac{dy}{dx}=0\\\\\\\dfrac{1}{\sqrt{y}}\dfrac{dy}{dx}=-\dfrac{1}{\sqrt{x}}\\\\\\\dfrac{dy}{dx}=-\dfrac{\sqrt{y}}{sqrt{x}}\\\\\\\boxed{\boxed{\dfrac{dy}{dx}=-\sqrt{\dfrac{y}{x}}}}$

b)

$e^{xy}+xy=0$

Derivando implicitamente em relação a x (a derivada de e^xy será encontrada pela regra da cadeia):

$e^{xy}\dfrac{d}{dx}(xy)+\dfrac{d}{dx}(xy)=0\\\\\\(e^{xy}+1)\cdot\dfrac{d}{dx}(xy)=0\\\\\\(e^{xy}+1)\cdot(y+x\dfrac{dy}{dx})=0\\\\\\ye^{xy}+xe^{xy}\dfrac{dy}{dx}+y+x\dfrac{dy}{dx}=0\\\\\\xe^{xy}\dfrac{dy}{dx}+x\dfrac{dx}{dy}=-ye^{xy}-y\\\\\\(xe^{xy}+x)\dfrac{dy}{dx}=-y(e^{xy}+1)\\\\\\\dfrac{dy}{dx}=-\dfrac{y(e^{xy}+1)}{xe^{x}+x}$

Colocando x em evidência no denominador:

$\dfrac{dy}{dx}=-\dfrac{y(e^{xy}+1)}{x(e^{xy+1})}\\\\\\\boxed{\boxed{\dfrac{dy}{dx}=-\dfrac{y}{x}}}$