Update linearApproximationWord1.tex

linearApproximation/exercises/linearApproximationWord1.tex

@@ -13,28 +13,13 @@
 surface area which results from this potential error in measurement.
 \begin{hint}
 The surface area of the cube, $S$, is a function of the edge length $x$:
-$S(x)=6 x^2$.
+$S(x)=\answer{6 x^2}$.
 \end{hint}
 \begin{hint}
-We have to compute the differential of $S$  at $x=30$.
-$\d S=S'(30)\d x$.
-\end{hint}
-\begin{hint}
-First, we have compute the derivative of $S$.
-$S'(x)=12 x$.
-\end{hint}
-\begin{hint}
-Now we  compute the the differential of $S$  at $x=30$.
-$\d S=12(30)\d x$.
-\end{hint}
-\begin{hint}
-We have to determine $\d x$.
-$\d x=\answer{1}$.
-\end{hint}
-\begin{hint}
-The potential error when we compute the surface area based on the measurement of the edge length  is then given by
-$\d S=12(30)\answer{1}$.
+Now we compute the differential of $S$  at $x=30$:
+$\d S=30\answer{12}\d x$, where $\dx= \answer{1}$
 \end{hint}
+
 \begin{prompt}
 	The potential error is $\pm \answer{12(30)} \textrm{ cm}^2$ 
 \end{prompt}