If x = log t and y + 1 = frac{1}{t},then e^{- | vedclass

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(A) Given $x = \log t$ and $y + 1 = \frac{1}{t}$.First,find the derivatives with respect to $t$:$\frac{dx}{dt} = \frac{1}{t}$ and $\frac{dy}{dt} = -\f

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(A) Given $x = \log t$ and $y + 1 = \frac{1}{t}$.First,find the derivatives with respect to $t$:$\frac{dx}{dt} = \frac{1}{t}$ and $\frac{dy}{dt} = -\frac{1}{t^{2}}$.Now,find $\frac{dx}{dy}$ using the chain rule:$\frac{dx}{dy} = \frac{dx/dt}{dy/dt} = \frac{1/t}{-1/t^{2}} = -t$.Next,find the second derivative $\frac{d^{2}x}{dy^{2}}$:$\frac{d^{2}x}{dy^{2}} = \frac{d}{dy}(-t) = \frac{d}{dt}(-t) \cdot \frac{dt}{dy} = (-1) \cdot \frac{1}{dy/dt} = (-1) \cdot \frac{1}{-1/t^{2}} = t^{2}$.Also,calculate $e^{-x}$:$e^{-x} = e^{-\log t} = e^{\log(t^{-1})} = \frac{1}{t}$.Finally,substitute these values into the expression $e^{-x} \frac{d^{2}x}{dy^{2}} + \frac{dx}{dy}$:$\left(\frac{1}{t}\right)(t^{2}) + (-t) = t - t = 0$.

If $x = \log t$ and $y + 1 = \frac{1}{t}$,then $e^{-x} \frac{d^{2} x}{d y^{2}} + \frac{d x}{d y} = $

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