#### Answer

$h'(x)=4x^{3}+4x$

#### Work Step by Step

$h(x)=(x^{2}+1)^{2}$
$h(x)=x^{4}+2x^{2}+1$
$h'(x)=4x^{4-1}+2x^{2-1}$
$h'(x)=4x^{3}+4x^{1}$
$h'(x)=4x^{3}+4x$
OR
$h(x)=(x^{2}+1)^{2}$
$h(x)=(x^{2}+1)(x^{2}+1)$
$u=x^{2}+1$
$u'=2x^{2-1}+0$
$u'=2x^{1}$
$u'=2x$
$v=u=x^{2}+1$
$v'=u'=2x$
$h'(x)=u'v+uv'$
$h'(x)=(2x)(x^{2}+1)+(x^{2}+1)(2x)$
$h'(x)=2x^{3}+2x+2x^{3}+2x$
$h'(x)=4x^{3}+4x$