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SOLUTION 20: Begin with the function
f(x)=1√1−x
and choose
x−values:0→h2
so that
Δx=h2−0=h2
The derivative of y=f(x) is
f′(x)=D{1√1−x}=D{(1−x)−1/2}=−12(1−x)−3/2⋅(−1)=12(1−x)3/2
The exact change of y−values is
Δy=f(h2)−f(0)
=1√1−(h2)−1√1−(0)
=1√1−h2−1√1
=1√1−h2−1
The Differential is
dy=f′(0) Δx
=12(1−(0))3/2⋅(h2)
=12(1)3/2⋅(h2)
=12(1)⋅(h2)
=12h2
Since h is "small" we will assume that
Δy≈dy ⟶
1√1−h2−1≈12h2 ⟶
1√1−h2≈1+12h2 ⟶
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