Chapter 5 - Graphs and the Derivative - 5.3 Higher Derivatives, Concavity, and the Second Derivative Test - 5.3 Exercises - Page 283: 8

$$f''\left( x \right) = \frac{{ - 2x\left( {{x^2} + 3} \right)}}{{{{\left( {1 - {x^2}} \right)}^3}}},f''\left( 0 \right) = 0,f''\left( 2 \right) = \frac{{28}}{{27}}$$

$$\eqalign{ & f\left( x \right) = \frac{{ - x}}{{1 - {x^2}}} \cr & {\text{find the derivative of }}f\left( x \right) \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {\frac{{ - x}}{{1 - {x^2}}}} \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^2}} \right)\frac{d}{{dx}}\left[ { - x} \right] + x\frac{d}{{dx}}\left[ {1 - {x^2}} \right]}}{{{{\left( {1 - {x^2}} \right)}^2}}} \cr & {\text{solve derivatives}} \cr & f'\left( x \right) = \frac{{\left( {1 - {x^2}} \right)\left( { - 1} \right) + x\left( { - 2x} \right)}}{{{{\left( {1 - {x^2}} \right)}^2}}} \cr & {\text{simplifying}} \cr & f'\left( x \right) = \frac{{{x^2} - 1 - 2{x^2}}}{{{{\left( {1 - {x^2}} \right)}^2}}} \cr & f'\left( x \right) = \frac{{ - 1 - {x^2}}}{{{{\left( {1 - {x^2}} \right)}^2}}} \cr & \cr & {\text{find the derivative of }}f'\left( x \right) \cr & f''\left( x \right) = \frac{d}{{dx}}\left[ {\frac{{ - 1 - {x^2}}}{{{{\left( {1 - {x^2}} \right)}^2}}}} \right] \cr & {\text{by using the quotient rule}} \cr & f''\left( x \right) = \frac{{{{\left( {1 - {x^2}} \right)}^2}\frac{d}{{dx}}\left[ { - 1 - {x^2}} \right] - \left( { - 1 - {x^2}} \right)\frac{d}{{dx}}\left[ {{{\left( {1 - {x^2}} \right)}^2}} \right]}}{{{{\left( {{{\left( {1 - {x^2}} \right)}^2}} \right)}^2}}} \cr & {\text{solve derivatives}}{\text{, use chain rule for }}\frac{d}{{dx}}\left[ {{{\left( {1 - {x^2}} \right)}^2}} \right] \cr & f''\left( x \right) = \frac{{{{\left( {1 - {x^2}} \right)}^2}\left( { - 2x} \right) - \left( { - 1 - {x^2}} \right)\left( 2 \right)\left( {1 - {x^2}} \right)\frac{d}{{dx}}\left[ {1 - {x^2}} \right]}}{{{{\left( {1 - {x^2}} \right)}^4}}} \cr & f''\left( x \right) = \frac{{{{\left( {1 - {x^2}} \right)}^2}\left( { - 2x} \right) - \left( { - 1 - {x^2}} \right)\left( 2 \right)\left( {1 - {x^2}} \right)\left( { - 2x} \right)}}{{{{\left( {1 - {x^2}} \right)}^4}}} \cr & {\text{factoring}} \cr & f''\left( x \right) = \frac{{ - 2x\left( {1 - {x^2}} \right)\left[ {1 - {x^2} - \left( { - 1 - {x^2}} \right)\left( 2 \right)} \right]}}{{{{\left( {1 - {x^2}} \right)}^4}}} \cr & f''\left( x \right) = \frac{{ - 2x\left[ {1 - {x^2} + 2 + 2{x^2}} \right]}}{{{{\left( {1 - {x^2}} \right)}^3}}} \cr & f''\left( x \right) = \frac{{ - 2x\left( {{x^2} + 3} \right)}}{{{{\left( {1 - {x^2}} \right)}^3}}} \cr & \cr & {\text{find }}f''\left( 0 \right){\text{ and }}f''\left( 2 \right) \cr & f''\left( 0 \right) = \frac{{ - 2\left( 0 \right)\left( {{0^2} + 3} \right)}}{{{{\left( {1 - {0^2}} \right)}^3}}} = 0 \cr & f''\left( 2 \right) = \frac{{ - 2\left( 2 \right)\left( {{2^2} + 3} \right)}}{{{{\left( {1 - {2^2}} \right)}^3}}} = \frac{{ - 28}}{{ - 27}} = \frac{{28}}{{27}} \cr} $$