## Calculus with Applications (10th Edition)

$$p'\left( t \right) = t{\left( {{t^2} + 1} \right)^{3/2}}\left( {7{t^2} + 1} \right)$$
\eqalign{ & p\left( t \right) = {t^2}{\left( {{t^2} + 1} \right)^{5/2}} \cr & {\text{differentiate both sides}} \cr & p'\left( t \right) = \frac{d}{{dt}}\left[ {{t^2}{{\left( {{t^2} + 1} \right)}^{5/2}}} \right] \cr & {\text{use the product rule}} \cr & p'\left( t \right) = {t^2}\frac{d}{{dt}}\left[ {{{\left( {{t^2} + 1} \right)}^{5/2}}} \right] + {\left( {{t^2} + 1} \right)^{5/2}}\frac{d}{{dt}}\left[ {{t^2}} \right] \cr & {\text{use the chain rule }}\frac{d}{{dx}}\left[ {g{{\left( x \right)}^n}} \right] = ng{\left( x \right)^{n - 1}}\frac{d}{{dx}}\left[ {g'\left( x \right)} \right]{\text{ then}} \cr & p'\left( t \right) = {t^2}\left( {\frac{5}{2}} \right){\left( {{t^2} + 1} \right)^{3/2}}\frac{d}{{dt}}\left[ {{t^2} + 1} \right] + {\left( {{t^2} + 1} \right)^{5/2}}\frac{d}{{dt}}\left[ {{t^2}} \right] \cr & {\text{find derivatives}} \cr & p'\left( t \right) = {t^2}\left( {\frac{5}{2}} \right){\left( {{t^2} + 1} \right)^{3/2}}\left( {2t} \right) + {\left( {{t^2} + 1} \right)^{5/2}}\left( {2t} \right) \cr & {\text{simplifying}} \cr & p'\left( t \right) = 5{t^3}{\left( {{t^2} + 1} \right)^{3/2}} + 2t{\left( {{t^2} + 1} \right)^{5/2}} \cr & p'\left( t \right) = t{\left( {{t^2} + 1} \right)^{3/2}}\left[ {5{t^2} + 2\left( {{t^2} + 1} \right)} \right] \cr & p'\left( t \right) = t{\left( {{t^2} + 1} \right)^{3/2}}\left( {7{t^2} + 1} \right) \cr}