Chapter 5 - Graphs and the Derivative - 5.3 Higher Derivatives, Concavity, and the Second Derivative Test - 5.3 Exercises - Page 283: 20

$$f'''\left( x \right) = 120{x^2} + 72x - 30,\,\,\,\,\,\,\,\,\,\,\,\,{f^{\left( 4 \right)}}\left( x \right) = 240x + 72$$

$$\eqalign{ & f\left( x \right) = 2{x^5} + 3{x^4} - 5{x^3} + 9x - 2 \cr & {\text{find the derivative of }}f\left( x \right) \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {2{x^5} + 3{x^4} - 5{x^3} + 9x - 2} \right] \cr & {\text{apply the power rule }}\frac{d}{{dx}}\left[ {{x^n}} \right] = {x^{n - 1}}{\text{ to each term}} \cr & f'\left( x \right) = 2\left( {5{x^4}} \right) + 3\left( {4{x^3}} \right) - 5\left( {3{x^2}} \right) + 9\left( 1 \right) - 0 \cr & f'\left( x \right) = 10{x^4} + 12{x^3} - 15{x^2} + 9 \cr & \cr & {\text{find the derivative of }}f'\left( x \right) \cr & f''\left( x \right) = \frac{d}{{dx}}\left[ {10{x^4} + 12{x^3} - 15{x^2} + 9} \right] \cr & {\text{use the power rule}} \cr & f''\left( x \right) = 10\left( {4{x^3}} \right) + 12\left( {3{x^2}} \right) - 15\left( {2x} \right) + 0 \cr & f''\left( x \right) = 40{x^3} + 36{x^2} - 30x \cr & \cr & {\text{find the derivative of }}f''\left( x \right) \cr & f'''\left( x \right) = \frac{d}{{dx}}\left[ {40{x^3} + 36{x^2} - 30x} \right] \cr & {\text{use the power rule}} \cr & f'''\left( x \right) = 40\left( {3{x^2}} \right) + 36\left( {2x} \right) - 30\left( 1 \right) \cr & f'''\left( x \right) = 120{x^2} + 72x - 30 \cr & \cr & {\text{find the derivative of }}f'''\left( x \right) \cr & {f^{\left( 4 \right)}}\left( x \right) = \frac{d}{{dx}}\left[ {120{x^2} + 72x - 30} \right] \cr & {\text{then}} \cr & {f^{\left( 4 \right)}}\left( x \right) = 120\left( {2x} \right) + 72\left( 1 \right) - 0 \cr & {f^{\left( 4 \right)}}\left( x \right) = 240x + 72 \cr} $$