Let h(t)=2t3.23t3.2\displaystyle {h}{\left({t}\right)}={2}{t}^{{{3.2}}}-{3}{t}^{{-{3.2}}}.
Then h(t)\displaystyle {h}'{\left({t}\right)} is  
h(3)\displaystyle {h}'{\left({3}\right)} is   ,

h(t)\displaystyle {h}{''}{\left({t}\right)} is  
and h(3)\displaystyle {h}{''}{\left({3}\right)} is