## Thomas' Calculus 13th Edition

$$y' = \frac{1}{{24}}{x^4},\,\,\,\,\,\,\,y'' = \frac{1}{6}{x^3},\,\,\,\,\,\,\,y''' = \frac{1}{2}{x^2},\,\,\,\,\,{y^{\left( 4 \right)}} = 2,\,\,\,\,\,{y^{\left( 5 \right)}} = 0$$
\eqalign{ & y = \frac{{{x^5}}}{{120}} \cr & {\text{Differentiate with respect to }}x \cr & y' = \frac{d}{{dx}}\left[ {\frac{{{x^5}}}{{120}}} \right] \cr & y' = \frac{1}{{120}}\frac{d}{{dx}}\left[ {{x^5}} \right] \cr & {\text{Use power rule }}\frac{d}{{dx}}\left[ {{x^n}} \right] = n{x^{n - 1}} \cr & y' = \frac{1}{{120}}\left( {5{x^4}} \right) \cr & y' = \frac{1}{{24}}{x^4} \cr & \cr & {\text{find the second derivative}} \cr & y'' = \frac{1}{{24}}\frac{d}{{dx}}\left[ {{x^4}} \right] \cr & y'' = \frac{1}{{24}}\left( {4{x^3}} \right) \cr & y'' = \frac{1}{6}{x^3} \cr & \cr & {\text{find the third derivative}} \cr & y''' = \frac{1}{6}\frac{d}{{dx}}\left[ {{x^3}} \right] \cr & y''' = \frac{1}{6}\left( {3{x^2}} \right) \cr & y''' = \frac{1}{2}{x^2} \cr & \cr & {\text{find the fourth derivative}} \cr & {y^{\left( 4 \right)}} = \frac{1}{2}\frac{d}{{dx}}\left[ {{x^2}} \right] \cr & {y^{\left( 4 \right)}} = \frac{1}{2}\left( {2x} \right) \cr & {y^{\left( 4 \right)}} = 2 \cr & \cr & {\text{find the fifth derivative}} \cr & {y^{\left( 5 \right)}} = \frac{d}{{dx}}\left[ 2 \right] \cr & {y^{\left( 5 \right)}} = 0 \cr}