a. By the chain rule,
[tex]\dfrac{dy}{dx} = \dfrac{dy}{dt}\cdot\dfrac{dt}{dx} = \dfrac{dy}{dt}\cdot\dfrac{1}{\frac{dx}{dt}}[/tex]
Given that [tex]y=e^t+e^{-t}[/tex] and [tex]x=e^{-t}[/tex], we have the derivatives
[tex]\dfrac{dy}{dt} = e^t -e^{-t}[/tex]
[tex]\dfrac{dx}{dt} = -e^{-t}[/tex]
and so
[tex]\dfrac{dy}{dx} = \dfrac{e^t - e^{-t}}{-e^{-t}} = 1-e^{2t}[/tex]
Set this equal to zero and solve for t :
[tex]1-e^{2t} = 0[/tex]
[tex]1 = e^{2t}[/tex]
[tex]\ln(1) = \ln\left(e^{2t}\right)[/tex]
[tex]0 = 2t \ln(e)[/tex]
[tex]0 = 2t[/tex]
[tex]t=0[/tex]
This value of t corresponds to x = e⁰ = 1 and y = e⁰ - 1/e⁰ = 1 - 1 = 0. So the only point on the curve where the derivative dy/dx is zero is (1, 0).
b. Compute the second derivative. Since dy/dx is a function of t, we'll momentarily replace it with f(t). By the chain rule,
[tex]\dfrac{d^2}{dx^2} = \dfrac d{dx} \dfrac{dy}{dx} = \dfrac{df}{dx} = \dfrac{df}{dt}\cdot\dfrac{dt}{dx} = \dfrac{df}{dt}\cdot\dfrac{1}{\frac{dx}{dt}}[/tex]
We have
[tex]\dfrac{df}{dt} = -2e^{2t}[/tex]
and we've already committed dx/dt. So
[tex]\dfrac{d^2}{dx^2} = \dfrac{-2e^{2t}}{-e^{-t}} = 2e^{3t} [/tex]
Substitute the first and second derivative into the differential equation:
[tex]\left(\dfrac{d^2y}{dx^2}\right)^2 + \dfrac{dy}{dx} - 1 = 0[/tex]
[tex]\left(2e^{3t}\right)^2 + (1-e^{2t}) - 1 = 0[/tex]
[tex]4e^{6t} -e^{2t}= 0[/tex]
[tex]e^{2t} (4e^{4t} - 1) = 0[/tex]
[tex]4e^{4t} - 1 = 0[/tex]
[tex]4e^{4t} = 1[/tex]
[tex]e^{4t} = \dfrac14[/tex]
[tex]\ln\left(e^{4t}\right) = \ln\left( \dfrac14\right)[/tex]
[tex]4t \ln(e) = -\ln(4)[/tex]
[tex]4t = -\ln(4)[/tex]
[tex]t = -\dfrac{\ln(4)}4[/tex]
