5. - / 1.66 points find \\( \\frac{d y}{d x} \\) and \\( \\frac{d^{2} y}{d x^{2}} \\). \\( x=e^{t}, \\quad…

Explanation:

Step1: Find (\frac{dx}{dt}) and (\frac{dy}{dt})

Using the derivative rules:

• For (x = e^{t}), (\frac{dx}{dt}=e^{t}) (since (\frac{d}{dt}(e^{at}) = ae^{at}), here (a = 1)).
• For (y=te^{-t}), use the product rule ((uv)^\prime=u^\prime v + uv^\prime) where (u = t), (u^\prime=1), (v = e^{-t}), (v^\prime=-e^{-t}). Then (\frac{dy}{dt}=e^{-t}-te^{-t}=(1 - t)e^{-t}).

Step2: Calculate (\frac{dy}{dx})

By the chain - rule (\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}). Substitute (\frac{dy}{dt}=(1 - t)e^{-t}) and (\frac{dx}{dt}=e^{t}) into the formula: (\frac{dy}{dx}=\frac{(1 - t)e^{-t}}{e^{t}}=(1 - t)e^{-2t}).

Step3: Find (\frac{d^{2}y}{dx^{2}})

First, find (\frac{d}{dt}(\frac{dy}{dx})). Let (u=(1 - t)) and (v = e^{-2t}). Using the product rule ((uv)^\prime=u^\prime v+uv^\prime), where (u^\prime=-1) and (v^\prime=-2e^{-2t}). (\frac{d}{dt}(\frac{dy}{dx})=-e^{-2t}-2(1 - t)e^{-2t}=(2t - 3)e^{-2t}). Then, by the formula (\frac{d^{2}y}{dx^{2}}=\frac{\frac{d}{dt}(\frac{dy}{dx})}{\frac{dx}{dt}}). Substitute (\frac{d}{dt}(\frac{dy}{dx})=(2t - 3)e^{-2t}) and (\frac{dx}{dt}=e^{t}) into the formula: (\frac{d^{2}y}{dx^{2}}=(2t - 3)e^{-3t}).

Step4: Determine when the curve is concave upward

The curve is concave upward when (\frac{d^{2}y}{dx^{2}}>0). Since (e^{-3t}>0) for all (t\in R) (because the exponential function (y = e^{ax}), (a=-3) has (y>0) for all (x)), we solve the inequality (2t-3>0). (2t-3>0\Rightarrow2t>3\Rightarrow t>\frac{3}{2}).

Answer:

(\frac{dy}{dx}=(1 - t)e^{-2t})

(\frac{d^{2}y}{dx^{2}}=(2t - 3)e^{-3t})

The curve is concave upward for (t\in(\frac{3}{2},\infty))