Given that \displaystyle y=6x\ {{e}^{{3-4x}}} and when \displaystyle x=\frac{3}{4}, there is increase in x of k % . Determine, in terms of k , the approximate percentage change in the value of y .
\displaystyle y=6x\ {{e}^{{3-4x}}}
\displaystyle \frac{{dy}}{{dx}}=6(x\frac{d}{{dx}}{{e}^{{3-4x}}}+{{e}^{{3-4x}}}\frac{{dx}}{{dx}})
\displaystyle \ \ \ \ \ =6(-4x\ {{e}^{{3-4x}}}+{{e}^{{3-4x}}})
\displaystyle \ \ \ \ \ =6{{e}^{{3-4x}}}(1-4x)
\displaystyle \text{when }x=\frac{3}{4},\ \ \frac{{dy}}{{dx}}=6\ {{e}^{0}}(-2)=-12
\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \delta x=\frac{3}{4}k\text{ }\!\!%\!\!\text{ } %
\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{3k}}{{400}}
\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \delta y\simeq (\frac{{dy}}{{dx}})\delta x
\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \simeq -12(\frac{{3k}}{{400}})
\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \simeq -\frac{{9k}}{{100}}
\displaystyle \text{when }x=\frac{3}{4},\ \ y=\frac{9}{2}
\displaystyle \text{Approximate percentage change in }y=\frac{{\delta y}}{y}\times 100\ %%
\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \simeq -\frac{{9k}}{{100}}\times \frac{2}{9}\times 100\ %%
\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \simeq -2k\ \text{ }\!\!%\!\!\text{ }%
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