Let $$y = {t^{10}} + 1$$   and $$x = {t^8} + 1,$$   then $$\frac{{{d^2}y}}{{d{x^2}}}$$  is equal to :

Solution :
$$\eqalign{ & {\text{We have, }}y = {t^{10}} + 1 \cr & \therefore \,\frac{{dy}}{{dt}} = 10{t^9} \cr & {\text{and }}x = {t^8} + 1 \Rightarrow \frac{{dx}}{{dt}} = 8{t^7} \cr & \therefore \,\frac{{dy}}{{dx}} = \frac{{\frac{{dy}}{{dt}}}}{{\frac{{dx}}{{dt}}}} = \frac{{10{t^9}}}{{8{t^7}}} = \frac{5}{4}\frac{{{t^{10}}}}{{{t^8}}} = \frac{{5\left( {y - 1} \right)}}{{4\left( {x - 1} \right)}}.....({\text{i}}) \cr & {\text{Differentiate (i) with respect to }}x,{\text{ we get}} \cr & \therefore \,\frac{{{d^2}y}}{{dx}} = \frac{5}{4}.\frac{{\left( {x - 1} \right).\frac{{dy}}{{dx}} - \left( {y - 1} \right)}}{{{{\left( {x - 1} \right)}^2}}} \cr & = \frac{5}{4}.\frac{1}{{\left( {x - 1} \right)}}\left[ {\frac{{dy}}{{dx}} - \frac{{\left( {y - 1} \right)}}{{\left( {x - 1} \right)}}} \right] \cr & = \frac{5}{4}.\frac{1}{{\left( {x - 1} \right)}}\left[ {\frac{5}{4}.\frac{{\left( {y - 1} \right)}}{{\left( {x - 1} \right)}} - \frac{{\left( {y - 1} \right)}}{{\left( {x - 1} \right)}}} \right]\,\,\,\,\,\,\,\,\left[ {{\text{Putting (i)}}} \right] \cr & = \frac{5}{4}.\frac{{\left( {y - 1} \right)}}{{{{\left( {x - 1} \right)}^2}}}\left( {\frac{5}{4} - 1} \right) \cr & = \frac{5}{{16}}.\frac{{{t^{10}}}}{{{{\left( {{t^8}} \right)}^2}}} \cr & = \frac{5}{{16{t^6}}} \cr} $$