Differentials Solution 15

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SOLUTION 15: Begin with the function

$f(x)= \displaystyle{ x \over 4 + x }$
and choose

$x-values: 0 \rightarrow h^2$
so that

$\Delta x = h^2 - 0 = h^2$
The derivative of

$\ y=f(x) \$ is

$f'(x)= \displaystyle{ (4+x)(1)-(x)(1) \over (4+x)^2 } = { 4 \over (4+x)^2 }$
The exact change of

$y-$ values is

$\Delta y = f(h^2) - f(0)$

$= \displaystyle{ (h^2) \over 4 + (h^2) } - \displaystyle{ (0) \over 4 + (0) }$

$= \displaystyle{ h^2 \over 4 + h^2 } - 0$

$= \displaystyle{ h^2 \over 4 + h^2 }$
The Differential is

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

$= \displaystyle{ 4 \over (4+(0))^2 } \cdot (h^2)$

$= \displaystyle{ 4 \over 16 } \cdot (h^2)$

$= \displaystyle{ 1 \over 4 } h^2$
Since

$h$ is "small" we will assume that

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

$\displaystyle{ h^2 \over 4 + h^2 } \approx \displaystyle{ 1 \over 4 } h^2$

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