Answer
The derivative is $\frac{(1-5t^2)(\sqrt t)}{2t}$
Work Step by Step
Product Rule $h'(t)=((u(t)(v(t))’=u'(t)v(t)+u(t)v’(t))$
$u(t)=\sqrt t ;u’(t)= \frac{1}{2\sqrt t}$
$v(t)=(1-t^2) ;v’(t)=-2t $ $h'(t)= (\frac{1}{2\sqrt t})(1-t^2)+(\sqrt t)((-2t)=$
$h'(t)= (\frac{1}{2\sqrt t})(1-t^2)+(\frac{2t}{2\sqrt t})((-2t)=$
$(\frac{1}{2\sqrt t})((1-t^2)+-4t^2)=$ $\frac{1-5t^2}{2\sqrt t}=\frac{(\sqrt t)(1-5t^2)}{2t}$
