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FGI2: Ableitungsregeln korrigiert

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Signed-off-by: Jim Martens <github@2martens.de>
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Jim Martens
2015-01-16 10:27:40 +01:00
parent 0fb0ec6ce6
commit 1d84978dd8
1 changed files with 3 additions and 3 deletions
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fgi2/Blatt12/Aufgabenblatt12.tex
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@ -127,7 +127,7 @@
Erste Ableitung:
\begin{alignat*}{2}
a \overset{a}{\rightarrow} \sqrt{} \;&&\; (A_0), x = a, v = a \\
a \cdot (b + (b + bb)) \overset{a}{\rightarrow} b + (b + bb) \;&&\; (T^{\sqrt{}}), x = a, v = a, y = b + (b + bb) \\
a \cdot (b + (b + bb)) \overset{a}{\rightarrow} b + (b + bb) \;&&\; (T.^{\sqrt{}}), x = a, v = a, y = b + (b + bb) \\
a \cdot (bb + a) \cdot b + a \cdot (b + (b + bb)) \overset{a}{\rightarrow} b + (b + bb) \;&&\; (T_{+L}), v = a, y = a \cdot (b + (b + bb))\\
&&\; x = a \cdot (bb + a) \cdot b, y' = b + (b + bb)
\end{alignat*}
@ -135,7 +135,7 @@
Zweite Ableitung:
\begin{alignat*}{2}
a \overset{a}{\rightarrow} \sqrt{} \;&&\; (A_0), x = a, v = a \\
a \cdot (bb + a) \cdot b \overset{a}{\rightarrow} (bb + a) \cdot b \;&&\; (T^{\sqrt{}}), x = a, v = a, y = (bb + a) \cdot b \\
a \cdot (bb + a) \cdot b \overset{a}{\rightarrow} (bb + a) \cdot b \;&&\; (T.^{\sqrt{}}), x = a, v = a, y = (bb + a) \cdot b \\
a \cdot (bb + a) \cdot b + a \cdot (b + (b + bb)) \overset{a}{\rightarrow} (bb + a) \cdot b \;&&\; (T_{+R}), x = a \cdot (bb + a) \cdot b, v = a \\
&&\; y = a \cdot (b + (b + bb)), x' = (bb + a) \cdot b
\end{alignat*}
@ -144,7 +144,7 @@
\begin{alignat*}{2}
a \overset{a}{\rightarrow} \sqrt{} \;&&\; (A_0), x = a, v = a \\
bb + a \overset{a}{\rightarrow} \sqrt{} \;&&\; (T^{\sqrt{}}_{+L}), y = a, v = a, x = bb \\
(bb + a) \cdot b \overset{a}{\rightarrow} b \;&&\; (T^{\sqrt{}}), x = (bb + a), v = a, y = b
(bb + a) \cdot b \overset{a}{\rightarrow} b \;&&\; (T.^{\sqrt{}}), x = (bb + a), v = a, y = b
\end{alignat*}
\subsection{}