Answer
$${\text{ }}\frac{1}{{\sqrt {1 - x} }} \approx 1 + \frac{1}{2}x$$
Work Step by Step
$$\eqalign{
& \frac{1}{{\sqrt {1 - x} }} \approx 1 + \frac{1}{2}x,{\text{ }}{x_0} = 0 \cr
& {\text{Let }}f\left( x \right) = \frac{1}{{\sqrt {1 - x} }} \cr
& {\text{ }}f\left( {{x_0}} \right) = f\left( 0 \right) = \frac{1}{{\sqrt {1 - 0} }} = 1 \cr
& f\left( x \right) = {\left( {1 - x} \right)^{ - 1/2}} \cr
& f'\left( x \right) = - \frac{1}{2}{\left( {1 - x} \right)^{ - 3/2}}\left( { - 1} \right) \cr
& f'\left( x \right) = \frac{1}{{2{{\left( {1 - x} \right)}^{3/2}}}} \cr
& f'\left( {{x_0}} \right) = f'\left( 0 \right) = \frac{1}{{2{{\left( {1 - 0} \right)}^{3/2}}}} = \frac{1}{2} \cr
& {\text{We can approximate values of }}f\left( x \right){\text{ by}} \cr
& {\text{ }}f\left( x \right) \approx f\left( {{x_0}} \right) + f'\left( {{x_0}} \right)\left( {x - {x_0}} \right) \cr
& {\text{then}} \cr
& {\text{ }}\frac{1}{{\sqrt {1 - x} }} \approx 1 + \frac{1}{2}\left( {x - 0} \right) \cr
& {\text{ }}\frac{1}{{\sqrt {1 - x} }} \approx 1 + \frac{1}{2}x \cr} $$
