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Update answer and latex in the hint of question 1 #108

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4 changes: 2 additions & 2 deletions week6/higher-order-derivatives.tex
Original file line number Diff line number Diff line change
Expand Up @@ -39,9 +39,9 @@
So $D^3f(0,0,0) = 2 dx\otimes dx \otimes dy+ 2 dx\otimes dy \otimes dx+2 dy\otimes dx \otimes dx$
\end{hint}
\begin{hint}
So $D^3f(0,0,0)(\verticalvector{1\\2},\verticalvector{3\\4},\verticalvector{0\\1}) =2( 2 \cdot 3 \cdot 1)+2( 1 \cdot 4 \cdot 0)+2(2 \cdot 3 \cdot 0) = 12$
So $D^3f(0,0,0)(\verticalvector{1\\2},\verticalvector{3\\4},\verticalvector{0\\1}) =2( 1 \cdot 3 \cdot 1)+2( 1 \cdot 4 \cdot 0)+2(2 \cdot 3 \cdot 0) = 6$
\end{hint}
$D^3f(0,0)(\verticalvector{1\\2},\verticalvector{3\\4},\verticalvector{0\\1})=$\answer{$12$}
$D^3f(0,0)(\verticalvector{1\\2},\verticalvector{3\\4},\verticalvector{0\\1})=$\answer{$6$}
\end{solution}
\end{question}

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