Differentials Solution 6

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SOLUTION 6: Begin with the function

$f(x)= e^{-x}$
and choose

$x-values: 0 \rightarrow 0.3$
so that

$\Delta x = 0.3-0 = 0.3$
The derivative of

$\ y=f(x) \$ is

$f'(x)= e^{-x} (-1) = -e^{-x}$
The exact change of

$y-$ values is

$\Delta y = f(0.3) - f(0)$

$= e^{-0.3} - e^{0}$

$= e^{-0.3} - 1$
The Differential is

$dy = f'(0) \ \Delta x$

$= -e^{0} \cdot (0.3)$

$= (-1) (0.3)$

$= -0.3$
We will assume that

$\Delta y \approx dy \ \ \ \ \longrightarrow$

$e^{-0.3} - 1 \approx -0.3 \ \ \ \ \longrightarrow$

$e^{-0.3} \approx 1 - 0.3 \ \ \ \ \longrightarrow$

$e^{-0.3} \approx 0.7$

NOTE: The number 0 was chosen for its proximity to 0.3 and for it's convenient exponential value. Check the accuracy of the final estimate using a CALCULATOR:

$e^{-0.3} \approx 0.7408$

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