Question:

<b>Implicit Higher Derivatives</b><br> At $x=0$: $e^y+0=e\Rightarrow y=1$.<br> Differentiate $e^y+xy=e$: $e^y y'+y+xy'=0\Rightarrow y'(e^y+x)=-y\Rightarrow y'=\dfrac{-y}{e^y+x}$.<br> At $(0,1)$: $y'=\dfrac{-1}{e+0}=-\dfrac{1}{e}$.<br> For $y''$: differentiate $y'(e^y+x)=-y$:<br> $y''(e^y+x)+y'(e^y y'+1)=-y'$<br> $y''(e+0)+(-1/e)(e\cdot(-1/e)+1)=-(-1/e)$<br> $y''\cdot e+(-1/e)(−1+1)=1/e$<br> $y''\cdot e+0=1/e\Rightarrow y''=\dfrac{1}{e^2}$.<br> Ordered pair: $\left(-\dfrac{1}{e},\dfrac{1}{e^2}\right)=$ option (1). <b>Answer: 1</b><br> <b>Key concept:</b> Find $y$ at the given point first (from original equation), then differentiate implicitly for $y'$ and $y''$.<br> <b>Trap:</b> Computing $y''$ by differentiating $y'=-y/(e^y+x)$ without using the implicit form — quotient rule gets messy; differentiating the implicit form is cleaner.