# Chapter 5 - Graphs and the Derivative - 5.3 Higher Derivatives, Concavity, and the Second Derivative Test - 5.3 Exercises - Page 283: 7

$$f''\left( x \right) = \frac{2}{{{{\left( {1 + x} \right)}^3}}},f''\left( 0 \right) = 2,f''\left( 2 \right) = \frac{2}{{27}}$$

#### Work Step by Step

\eqalign{ & f\left( x \right) = \frac{{{x^2}}}{{1 + x}} \cr & {\text{find the derivative of }}f\left( x \right) \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {\frac{{{x^2}}}{{1 + x}}} \right] \cr & {\text{by using the quotient rule }}\frac{d}{{dx}}\left[ {\frac{{u\left( x \right)}}{{v\left( x \right)}}} \right] = \frac{{v\left( x \right) \cdot u'\left( x \right) - u\left( x \right)v'\left( x \right)}}{{{{\left[ {v\left( x \right)} \right]}^2}}} \cr & f'\left( x \right) = \frac{{\left( {1 + x} \right)\frac{d}{{dx}}\left[ {{x^2}} \right] - {x^2}\frac{d}{{dx}}\left[ {1 + x} \right]}}{{{{\left( {1 + x} \right)}^2}}} \cr & {\text{solve derivatives}} \cr & f'\left( x \right) = \frac{{\left( {1 + x} \right)\left( {2x} \right) - {x^2}\left( 1 \right)}}{{{{\left( {1 + x} \right)}^2}}} \cr & {\text{simplifying}} \cr & f'\left( x \right) = \frac{{2x + 2{x^2} - {x^2}}}{{1 + 2x + {x^2}}} \cr & f'\left( x \right) = \frac{{2x + {x^2}}}{{1 + 2x + {x^2}}} \cr & \cr & {\text{find the derivative of }}f'\left( x \right) \cr & f''\left( x \right) = \frac{d}{{dx}}\left[ {\frac{{2x + {x^2}}}{{1 + 2x + {x^2}}}} \right] \cr & {\text{by using the quotient rule}} \cr & f''\left( x \right) = \frac{{\left( {1 + 2x + {x^2}} \right)\frac{d}{{dx}}\left[ {2x + {x^2}} \right] - \left( {2x + {x^2}} \right)\frac{d}{{dx}}\left[ {1 + 2x + {x^2}} \right]}}{{{{\left( {1 + 2x + {x^2}} \right)}^2}}} \cr & {\text{solve derivatives}} \cr & f''\left( x \right) = \frac{{\left( {1 + 2x + {x^2}} \right)\left( {2 + 2x} \right) - \left( {2x + {x^2}} \right)\left( {2 + 2x} \right)}}{{{{\left( {1 + 2x + {x^2}} \right)}^2}}} \cr & {\text{simplifying}} \cr & f''\left( x \right) = \frac{{\left( {2 + 2x} \right)\left[ {1 + 2x + {x^2} - 2x - {x^2}} \right]}}{{{{\left( {1 + 2x + {x^2}} \right)}^2}}} \cr & f''\left( x \right) = \frac{{\left( {2 + 2x} \right)\left( 1 \right)}}{{{{\left( {1 + x} \right)}^4}}} \cr & f''\left( x \right) = \frac{{2\left( {1 + x} \right)\left( 1 \right)}}{{{{\left( {1 + x} \right)}^4}}} \cr & f''\left( x \right) = \frac{2}{{{{\left( {1 + x} \right)}^3}}} \cr & \cr & {\text{find }}f''\left( 0 \right){\text{ and }}f''\left( 2 \right) \cr & f''\left( 0 \right) = \frac{2}{{{{\left( {1 + 0} \right)}^3}}} = 2 \cr & f''\left( 2 \right) = \frac{2}{{{{\left( {1 + 2} \right)}^3}}} = \frac{2}{{27}} \cr}

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