Answer
$f'\left( x \right) = - \frac{1}{{2{{\left( {1 + x} \right)}^{3/2}}}}$
$D_f=(-1,\infty)$
$D_{f'}=(-1,\infty)$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \frac{1}{{\sqrt {1 + x} }} \cr
& {\text{The domain of the function is }}\left( { - 1,\infty } \right) \cr
& {\text{Using the definition of derivative}}{\text{. }} \cr
& f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h} \cr
& {\text{Therefore}}{\text{,}} \cr
& f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{\frac{1}{{\sqrt {1 + \left( {x + h} \right)} }} - \frac{1}{{\sqrt {1 + x} }}}}{h} \cr
& f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{\sqrt {1 + x} - \sqrt {1 + \left( {x + h} \right)} }}{{h\sqrt {1 + x} \sqrt {1 + \left( {x + h} \right)} }} \cr
& {\text{Rationalizing the numerator}} \cr
& f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{\sqrt {1 + x} - \sqrt {1 + \left( {x + h} \right)} }}{{h\sqrt {1 + x} \sqrt {1 + \left( {x + h} \right)} }} \times \frac{{\sqrt {1 + x} + \sqrt {1 + \left( {x + h} \right)} }}{{\sqrt {1 + x} + \sqrt {1 + \left( {x + h} \right)} }} \cr
& f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{{{\left( {\sqrt {1 + x} } \right)}^2} - {{\left( {\sqrt {1 + \left( {x + h} \right)} } \right)}^2}}}{{h\left( {\sqrt {1 + x} \sqrt {1 + \left( {x + h} \right)} } \right)\left( {\sqrt {1 + x} + \sqrt {1 + \left( {x + h} \right)} } \right)}} \cr
& f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{1 + x - 1 - \left( {x + h} \right)}}{{h\left( {\sqrt {1 + x} \sqrt {1 + \left( {x + h} \right)} } \right)\left( {\sqrt {1 + x} + \sqrt {1 + \left( {x + h} \right)} } \right)}} \cr
& f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{1 + x - 1 - x - h}}{{h\left( {\sqrt {1 + x} \sqrt {1 + \left( {x + h} \right)} } \right)\left( {\sqrt {1 + x} + \sqrt {1 + \left( {x + h} \right)} } \right)}} \cr
& f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{ - h}}{{h\left( {\sqrt {1 + x} \sqrt {1 + \left( {x + h} \right)} } \right)\left( {\sqrt {1 + x} + \sqrt {1 + \left( {x + h} \right)} } \right)}} \cr
& f'\left( x \right) = - \mathop {\lim }\limits_{h \to 0} \frac{1}{{\left( {\sqrt {1 + x} \sqrt {1 + \left( {x + h} \right)} } \right)\left( {\sqrt {1 + x} + \sqrt {1 + \left( {x + h} \right)} } \right)}} \cr
& {\text{Evaluate the limit when }}h \to 0 \cr
& f'\left( x \right) = - \mathop {\lim }\limits_{h \to 0} \frac{1}{{\left( {\sqrt {1 + x} \sqrt {1 + x} } \right)\left( {\sqrt {1 + x} + \sqrt {1 + x} } \right)}} \cr
& f'\left( x \right) = - \mathop {\lim }\limits_{h \to 0} \frac{1}{{\left( {1 + x} \right)\left( {2\sqrt {1 + x} } \right)}} \cr
& f'\left( x \right) = - \mathop {\lim }\limits_{h \to 0} \frac{1}{{2{{\left( {1 + x} \right)}^{3/2}}}} \cr
& {\text{The domain of the derivative of the function is is }}\left( { - 1,\infty } \right) \cr} $$
