Question

A equação da reta tangente à curva definida pela equação x^2y - xy^3 + 2 = y no ponto de abscissa x = 0 é :
a) 8x - y = -2
b) 8x + y = 2
c) x - 8y = 2
d) x + 8y = 2
e) x + y = 2

Answer 1

$\boxed{\begin{array}{l}\rm x^2y-xy^3+2=y\\\rm0^2y-0y^3+2=y\\\rm y=2\longrightarrow P(0,2)\\\rm vamos~achar~a~derivada\\\rm usando~diferenciac_{\!\!,}\tilde ao~impl\acute icita.\\\rm2xy+x^2\dfrac{dy}{dx}-\bigg(y^3+3xy^2\dfrac{dy}{dx}\bigg)+0=\dfrac{dy}{dx}\\\\\rm\dfrac{dy}{dx}=2xy+x^2\dfrac{dy}{dx}-y^3-3xy^2\dfrac{dy}{dx}\end{array}}$

$\boxed{\begin{array}{l}\rm\dfrac{dy}{dx}-x^2\dfrac{dy}{dx}+3xy^2\dfrac{dy}{dx}=2xy-y^3\\\\\rm\dfrac{dy}{dx}(1-x^2+3xy^2)=2xy-y^3\\\\\rm\dfrac{dy}{dx}=\dfrac{2xy-y^3}{1-x^2+3xy^2}\\\\\rm\dfrac{dy}{dx}\bigg|_{x=0~y=2}=\dfrac{2\cdot0\cdot2-2^3}{1-0^2+3\cdot0\cdot2^2}\\\\\rm\dfrac{dy}{dx}\bigg|_{x=0~y=2}=-\dfrac{8}{1}=-8\\\\\rm y=y_0+\dfrac{dy}{dx}\bigg|_{x=0~y=2}\cdot(x-x_0)\\\\\rm y=2-8(x-0)\\\rm y=2-8x\\\rm 8x+y=2\\\huge\boxed{\boxed{\boxed{\boxed{\rm\dagger\red{\maltese}~\blue{alternativa~b}}}}}\end{array}}$