We want to apply the chain rule. Set u = x − 1 {\displaystyle u={\sqrt {x}}-1} .
g ( u ) = u − 1 g ′ ( u ) = 1 2 u u ( x ) = x − 1 u ′ ( x ) = 1 2 x {\displaystyle {\begin{aligned}g(u)&={\sqrt {u}}-1&g'(u)&={\frac {1}{2{\sqrt {u}}}}\\u(x)&={\sqrt {x}}-1&u'(x)&={\frac {1}{2{\sqrt {x}}}}\end{aligned}}}
Combining, we get:
g ′ ( x ) = 1 2 x ⋅ 1 2 x − 1 = 1 4 x ⋅ x − 1 = 1 4 x ( x − 1 ) {\displaystyle {\begin{aligned}g'(x)&={\frac {1}{2{\sqrt {x}}}}\cdot {\frac {1}{2{\sqrt {{\sqrt {x}}-1}}}}\\&={\frac {1}{4{\sqrt {x}}\cdot {\sqrt {{\sqrt {x}}-1}}}}\\&={\frac {1}{4{\sqrt {x\left({\sqrt {x}}-1\right)}}}}\end{aligned}}}
.
