Question

derive até a segunda ordem t2+5t-1/t2

Answer 1

$\textbf{Calculando a derivada primeira:}\\\\\\ \mathrm{y=\dfrac{t^2+5t-1}{t^2}\ \to\ \dfrac{dy}{dt}=\dfrac{\frac{d}{dt}(t^2+5t-1).t^2-\frac{d}{dt}(t^2)(t^2+5t-1)}{\big(t^2\big)^2}}\\\\\\ \mathrm{\dfrac{dy}{dt}=\dfrac{(2t+5).t^2-2t(t^2+5t-1)}{t^4}=\dfrac{(2t+5)t^2}{t^4}-\dfrac{2t(t^2+5t-1)}{t^4}}\\\\\\ \mathrm{\dfrac{dy}{dt}=\dfrac{t(2t+5)}{t^3}-\dfrac{2(t^2+5t-1)}{t^3}=\dfrac{2t^2+5t-2t^2-10t+2}{t^3}}\\\\\\ \mathrm{\dfrac{dy}{dt}=\dfrac{-5t+2}{t^3}\ \to\ \mathbf{\dfrac{dy}{dt}=-\dfrac{5t-2}{t^3}}}$

$\textbf{Calculando a derivada segunda:}\\\\\\ \mathrm{\dfrac{d^2y}{dt^2}=-\dfrac{dy}{dt}\bigg(\dfrac{5t-2}{t^3}\bigg)=-\dfrac{\frac{d}{dt}(5t-2).t^3-\frac{d}{dt}(t^3)(5t-2)}{\big(t^3\big)^2}}\\\\\\ \mathrm{\dfrac{d^2y}{dt^2}=-\dfrac{5t^3-3t^2(5t-2)}{t^6}=\dfrac{3t^2(5t-2)}{t^6}-\dfrac{5t^3}{t^6}}\\\\\\ \mathrm{\dfrac{d^2y}{dt^2}=\dfrac{3(5t-2)}{t^4}-\dfrac{5t}{t^4}=\dfrac{15t-6-5t}{t^4}\ \to\ \mathbf{\dfrac{d^2y}{dt^2}=\dfrac{10t-6}{t^4}}}$