Answer to Question #271911 in Calculus for Angel Nodado

Question

Find the first and second derivatives of the following and simplify whenever possible:

x = t ^ 2 * e ^ t y = t In t

Answer 1

$$\begin{matrix} x=t^2e^t \\ y=t\ln t \end{matrix}$$

$$\dfrac{dx}{dt}=2te^t+t^2e^t$$

$$\dfrac{dy}{dt}=\ln t+t(\dfrac{1}{t})=\ln t+1$$

$$\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}=\dfrac{\ln t+1}{2te^t+t^2e^t}$$

$$\dfrac{d}{dt}(\dfrac{dy}{dx})=\dfrac{\dfrac{1}{t}(2te^t+t^2e^t)-(2e^t+4te^t+t^2e^t)(\ln t+1)}{(2te^t+t^2e^t)^2}$$

$$=\dfrac{2e^t+te^t-(2e^t+4te^t+t^2e^t)\ln t-2e^t-4te^t-t^2e^t}{(2te^t+t^2e^t)^2}$$

$$=-\dfrac{3te^t+t^2e^t+(2e^t+4te^t+t^2e^t)\ln t}{(2te^t+t^2e^t)^2}$$

$$\dfrac{d^2y}{dx^2}=\dfrac{\dfrac{d}{dt}(\dfrac{dy}{dx})}{dx/dt}$$

$$=-\dfrac{3te^t+t^2e^t+(2e^t+4te^t+t^2e^t)\ln t}{(2te^t+t^2e^t)^3}$$