Answer
$$\eqalign{
& {\bf{r}}''\left( t \right) = \left\langle {2{{\sec }^2}t\tan t,2{t^{ - 3}},{{\left( {t + 1} \right)}^{ - 2}}} \right\rangle \cr
& {\bf{r}}'''\left( t \right) = \left\langle {2{{\sec }^4}t + 4{{\sec }^2}t{{\tan }^2}t, - 6{t^{ - 4}}, - 2{{\left( {t + 1} \right)}^{ - 3}}} \right\rangle \cr} $$
Work Step by Step
$$\eqalign{
& {\bf{r}}\left( t \right) = \tan t{\bf{i}} + \left( {t + \frac{1}{t}} \right){\bf{j}} - \ln \left( {t + 1} \right){\bf{k}} \cr
& {\text{Differentiate}} \cr
& {\bf{r}}'\left( t \right) = {\sec ^2}t{\bf{i}} + \left( {1 - \frac{1}{{{t^2}}}} \right){\bf{j}} - \frac{1}{{t + 1}}{\bf{k}} \cr
& {\bf{r}}'\left( t \right) = \left\langle {{{\sec }^2}t,1 - {t^{ - 2}}, - \frac{1}{{t + 1}}} \right\rangle \cr
& \cr
& {\bf{r}}''\left( t \right) = \frac{d}{{dt}}\left[ {{{\sec }^2}t} \right]{\bf{i}} + \frac{d}{{dt}}\left( {1 - \frac{1}{{{t^2}}}} \right){\bf{j}} - \frac{d}{{dt}}\left[ {\frac{1}{{t + 1}}} \right]{\bf{k}} \cr
& {\bf{r}}''\left( t \right) = 2\sec t\sec t\tan t{\bf{i}} + \frac{2}{{{t^3}}}{\bf{j}} + \frac{1}{{{{\left( {t + 1} \right)}^2}}}{\bf{k}} \cr
& {\bf{r}}''\left( t \right) = \left\langle {2{{\sec }^2}t\tan t,2{t^{ - 3}},{{\left( {t + 1} \right)}^{ - 2}}} \right\rangle \cr
& \cr
& {\bf{r}}'''\left( t \right) = \frac{d}{{dt}}\left[ {2{{\sec }^2}t\tan t} \right]{\bf{i}} + \frac{d}{{dt}}\left[ {2{t^{ - 3}}} \right]{\bf{j}} + \frac{d}{{dt}}\left[ {{{\left( {t + 1} \right)}^{ - 2}}} \right]{\bf{k}} \cr
& {\bf{r}}'''\left( t \right) = \left( {2{{\sec }^2}t{{\sec }^2}t + 4{{\sec }^2}t{{\tan }^2}t} \right){\bf{i}} + \left( { - 6{t^{ - 4}}} \right){\bf{j}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\, + \left( { - 2{{\left( {t + 1} \right)}^{ - 3}}} \right){\bf{k}} \cr
& {\bf{r}}'''\left( t \right) = \left\langle {2{{\sec }^4}t + 4{{\sec }^2}t{{\tan }^2}t, - 6{t^{ - 4}}, - 2{{\left( {t + 1} \right)}^{ - 3}}} \right\rangle \cr} $$
