Question

Dado y=(x²+1)^1/2, tem -se y'

Answer 1

derivando usando a regra da cadeia
$(u^{N} )' = N*U^{(N-1)} * U'$

aplicando isso:
$y=(x^2+1)^{ \frac{1}{2} }\\\\ y'= \frac{1}{2}*(x^2+1)^{( \frac{1}{2}-1) } * (x^2+1)'\\\\y'= \frac{1}{2}*(x^2+1)^{ \frac{-1}{2} }*(2x^{x-1}+0)\\\\y'= \frac{1}{\not 2}*(x^2+1)^{- \frac{1}{2} } *(\not 2x)\\\\y'=(x^2+1)^{-\frac{1}{2}}$