Question

Cálculo I simples (imagem)

Answer 1

Explicação passo-a-passo:

Temos que $y(x)=\dfrac{x+1}{x-1}$.

Calculando a primeira derivada com a Regra do Quociente:

$\dfrac{dy}{dx}=\dfrac{\frac{d}{dx}(x+1).(x-1)-(x+1).\frac{d}{dx}(x-1) }{(x-1)^{2} } \\\\\dfrac{dy}{dx}=\dfrac{1.(x-1)-(x+1).1 }{(x-1)^{2} } \\\\\dfrac{dy}{dx}=\dfrac{x-1-x-1 }{(x-1)^{2} } \\\\\boxed{\dfrac{dy}{dx}=\dfrac{-2 }{(x-1)^{2} } }$

Calculando a segunda derivada com a Regra do Quociente:

$\dfrac{d^{2} y}{dx^{2} }=\dfrac{\frac{d}{dx} (-2).(x-1)^{2} -(-2).\frac{d}{dx}(x-1)^{2} }{((x-1)^{2})^{2} }\\\\\dfrac{d^{2} y}{dx^{2} }=\dfrac{0.(x-1)^{2}+2.\frac{d}{dx}(x^{2}-2x +1) }{(x-1)^{4} }\\\\\dfrac{d^{2} y}{dx^{2} }=\dfrac{2.(2x-2) }{(x-1)^{4} }\\\\\dfrac{d^{2} y}{dx^{2} }=\dfrac{4.(x-1) }{(x-1)^{4} }\\\\\boxed{\dfrac{d^{2} y}{dx^{2} }=\dfrac{4 }{(x-1)^{3} }}$

Verificando a igualdada:

$(1-x)\dfrac{d^{2} y}{dx^{2} } =2\dfrac{dy}{dx} \\\\\text Multiplicando\ tudo\ por\ -1:\\-(1-x)\dfrac{d^{2} y}{dx^{2} } =-2\dfrac{dy}{dx}\\\\(x-1)\dfrac{d^{2} y}{dx^{2} } =-2\dfrac{dy}{dx}\\\\(x-1).\dfrac{4}{(x-1)^{3} }=(-2).\dfrac{-2}{(x-1)^{2}} \\\\\boxed{\boxed{\dfrac{4}{(x-1)^{2} }=\dfrac{4}{(x-1)^{2}}}}$

Essa igualdade é Verdadeira.