Question #271911

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



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

1
Expert's answer
2021-11-29T13:15:56-0500
x=t2ety=tlnt\begin{matrix} x=t^2e^t \\ y=t\ln t \end{matrix}

dxdt=2tet+t2et\dfrac{dx}{dt}=2te^t+t^2e^t

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

dydx=dy/dtdx/dt=lnt+12tet+t2et\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}=\dfrac{\ln t+1}{2te^t+t^2e^t}

ddt(dydx)=1t(2tet+t2et)(2et+4tet+t2et)(lnt+1)(2tet+t2et)2\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}

=2et+tet(2et+4tet+t2et)lnt2et4tett2et(2tet+t2et)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}

=3tet+t2et+(2et+4tet+t2et)lnt(2tet+t2et)2=-\dfrac{3te^t+t^2e^t+(2e^t+4te^t+t^2e^t)\ln t}{(2te^t+t^2e^t)^2}

d2ydx2=ddt(dydx)dx/dt\dfrac{d^2y}{dx^2}=\dfrac{\dfrac{d}{dt}(\dfrac{dy}{dx})}{dx/dt}

=3tet+t2et+(2et+4tet+t2et)lnt(2tet+t2et)3=-\dfrac{3te^t+t^2e^t+(2e^t+4te^t+t^2e^t)\ln t}{(2te^t+t^2e^t)^3}


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