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If $x=\log t, y+1=\frac{1}{t}, \quad$ then $e^{-x} \frac{d^{2} x}{d y^{2}}+\frac{d x}{d y}=$
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Given $x=\log t$ and $y+1=\frac{1}{t}$
$\therefore \frac{d x}{d t}=\frac{1}{t}$ and $\frac{d y}{d t}=\frac{-1}{t^{2}}$
$\therefore \frac{d x}{d y}=\frac{1}{t} \times\left(-t^{2}\right)=-t$
$\therefore \frac{d^{2} x}{d y^{2}}=\frac{d}{d t}\left(\frac{d x}{d y}\right) \times \frac{d t}{d y}=\frac{d}{d t}(-t) \times \frac{1}{\left(\frac{d y}{d t}\right)}=\frac{(-1)}{\left(\frac{-1}{t^{2}}\right)}=t^{2}$
$e^{-x}=e^{-\log t}=e^{\log (t)^{-1}}=\frac{1}{t}$
Thus $e^{-x} \frac{d^{2} y}{d x^{2}}+\frac{d x}{d y}$
$=\left(\frac{1}{t}\right)\left(t^{2}\right)+(-t)=t-t=0$
$\therefore \frac{d x}{d t}=\frac{1}{t}$ and $\frac{d y}{d t}=\frac{-1}{t^{2}}$
$\therefore \frac{d x}{d y}=\frac{1}{t} \times\left(-t^{2}\right)=-t$
$\therefore \frac{d^{2} x}{d y^{2}}=\frac{d}{d t}\left(\frac{d x}{d y}\right) \times \frac{d t}{d y}=\frac{d}{d t}(-t) \times \frac{1}{\left(\frac{d y}{d t}\right)}=\frac{(-1)}{\left(\frac{-1}{t^{2}}\right)}=t^{2}$
$e^{-x}=e^{-\log t}=e^{\log (t)^{-1}}=\frac{1}{t}$
Thus $e^{-x} \frac{d^{2} y}{d x^{2}}+\frac{d x}{d y}$
$=\left(\frac{1}{t}\right)\left(t^{2}\right)+(-t)=t-t=0$
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