Find dydx and d2ydx2.x=et,y=te−t. For which values of t is the curve concave upward?

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crocolylec · Accepted Answer

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Jeffrey Jordon · Accepted Answer

Answer:y′=1−ln⁡xx2andy″=2ln⁡(x)−3x3The curve is concave upward in the interval t∈(32,∞)Explanation:To determine the intervals of concavity, we apply the second derivative test to the function y=y(x), i.e. determine y″=d2ydx2To find the first and second derivative dydx and d2ydx2as followsy=te−t=tet=ln⁡(x)xwhere x=et⇒ln⁡(x)=ln⁡(et)=tln⁡(e)=tThus, the parametric equations x=et and y=te−t given equation (1) are represented by the cartesian equation (1) are represented by the cartesian equationy=ln⁡xxwhere x=et⇒ln⁡(x)=ln⁡(et)=tln⁡(e)=tThus, the parametric equations x=et and y=te−t given in equation are represented by the cartesian equationy=ln⁡(x)xTo find the first derivative to the cartesian function we apply the quotient rule of derivatives as followsddx[u(x)v(x)]=u′v−v′u[v(x)]2Thus,y′=dydx=[ln⁡(x)]′x−[x]′ln⁡(x)[x]2=[1x]x−[1]ln⁡(x)[x]2=1−ln⁡(x)x2Again, we apply the quotient rule of derivatives to the function y′=1−ln⁡(x)x2 as followsy″=d2ydx2=[1−ln⁡(x)]′x2−[x2]′[1−ln⁡(x)][x2]2=[0−1x]x2−[2x][1−ln⁡(x)]x4=−x−2x[1−ln⁡(x)]x4=x(−1−2[1−ln⁡(x)])x4=−1−2[1−ln⁡(x)]x3=−1−2+2ln⁡(x)x3=2ln⁡(x)−3x3The relation the concavity of the graph of the function to its derivative is defined as followsThe graph of the function f(x) is concave upward if f″(x)&amp;gt;0 and concave downward if f″(x)&amp;lt;0Thus, from concavity theorem, we obtainThe graph of the function y=f(x) is concave upward wheny″=2ln⁡(x)−3x3&amp;gt;0Thus, the inequality implies that2ln⁡(x)−3&amp;gt;0 and

Find \frac{dy}{dx} and \frac{d^2y}{dx^2}.x=e^t,y=te^{-t}. For which values of t is the curve concave upward?

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