Approximation

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{ }$%