Differentials Solution 20

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SOLUTION 20: Begin with the function

$f(x)= \displaystyle{ 1 \over \sqrt{1-x} }$
and choose

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

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

$\ y=f(x) \$ is

$f'(x)= D \Big\{ \displaystyle{ 1 \over \sqrt{1-x} } \Big\} = D\{(1-x)^{-1/2} \} = { -1 \over 2 } (1-x)^{-3/2} \cdot (-1) = { 1 \over 2(1-x)^{3/2} }$
The exact change of

$y-$ values is

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

$= \displaystyle{ 1 \over \sqrt{1-(h^2)} } - \displaystyle{ 1 \over \sqrt{1-(0)} }$

$= \displaystyle{ 1 \over \sqrt{1-h^2} } - \displaystyle{ 1 \over \sqrt{1} }$

$= \displaystyle{ 1 \over \sqrt{1-h^2} } - 1$
The Differential is

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

$= \displaystyle { 1 \over 2(1-(0))^{3/2} } \cdot (h^2)$

$= \displaystyle { 1 \over 2(1)^{3/2} } \cdot (h^2)$

$= \displaystyle { 1 \over 2(1) } \cdot (h^2)$

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

$h$ is "small" we will assume that

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

$\displaystyle{ 1 \over \sqrt{1-h^2} } - 1 \approx \displaystyle{ 1 \over 2 }h^2 \ \ \ \ \longrightarrow$

$\displaystyle{ 1 \over \sqrt{1-h^2} } \approx 1+ \displaystyle{ 1 \over 2 }h^2 \ \ \ \ \longrightarrow$

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