Chapter 13 - Partial Derivatives - 13.3 Partial Derivatives - Exercises Set 13.3 - Page 937: 45

$$\eqalign{ & {f_x}\left( {x,y,z} \right) = - \frac{{{y^2}{z^3}}}{{\left( {{x^2}{y^4}{z^6} + 1} \right)}} \cr & {f_y}\left( {x,y,z} \right) = - \frac{{2xy{z^3}}}{{\left( {{x^2}{y^4}{z^6} + 1} \right)}} \cr & {f_z}\left( {x,y,z} \right) = - \frac{{3x{y^2}{z^2}}}{{\left( {{x^2}{y^4}{z^6} + 1} \right)}} \cr} $$

$$\eqalign{ & f\left( {x,y,z} \right) = {\tan ^{ - 1}}\left( {\frac{1}{{x{y^2}{z^3}}}} \right) \cr & {\text{Calculate }}{f_x}\left( {x,y,z} \right) \cr & {f_x}\left( {x,y,z} \right) = \frac{\partial }{{\partial x}}\left[ {{{\tan }^{ - 1}}\left( {\frac{1}{{x{y^2}{z^3}}}} \right)} \right] \cr & {f_x}\left( {x,y,z} \right) = \frac{{\frac{\partial }{{\partial x}}\left[ {\frac{1}{{x{y^2}{z^3}}}} \right]}}{{1 + {{\left( {1/x{y^2}{z^3}} \right)}^2}}} \cr & {f_x}\left( {x,y,z} \right) = \frac{{ - \frac{1}{{{x^2}{y^2}{z^3}}}}}{{1 + 1/{x^2}{y^4}{z^6}}} \cr & {f_x}\left( {x,y,z} \right) = \frac{{ - \frac{1}{{{x^2}{y^2}{z^3}}}}}{{\left( {{x^2}{y^4}{z^6} + 1} \right)/{x^2}{y^4}{z^6}}} \cr & {f_x}\left( {x,y,z} \right) = - \frac{{{x^2}{y^4}{z^6}}}{{{x^2}{y^2}{z^3}\left( {{x^2}{y^4}{z^6} + 1} \right)}} \cr & {f_x}\left( {x,y,z} \right) = - \frac{{{y^2}{z^3}}}{{\left( {{x^2}{y^4}{z^6} + 1} \right)}} \cr & \cr & {\text{Calculate }}{f_y}\left( {x,y,z} \right) \cr & {f_y}\left( {x,y,z} \right) = \frac{\partial }{{\partial y}}\left[ {{{\tan }^{ - 1}}\left( {\frac{1}{{x{y^2}{z^3}}}} \right)} \right] \cr & {f_y}\left( {x,y,z} \right) = \frac{{\frac{\partial }{{\partial y}}\left[ {\frac{1}{{x{y^2}{z^3}}}} \right]}}{{1 + {{\left( {1/x{y^2}{z^3}} \right)}^2}}} \cr & {f_y}\left( {x,y,z} \right) = \frac{{ - \frac{2}{{x{y^3}{z^3}}}}}{{1 + 1/{x^2}{y^4}{z^6}}} \cr & {f_y}\left( {x,y,z} \right) = \frac{{ - \frac{2}{{x{y^3}{z^3}}}}}{{\left( {{x^2}{y^4}{z^6} + 1} \right)/{x^2}{y^4}{z^6}}} \cr & {f_y}\left( {x,y,z} \right) = - \frac{{2{x^2}{y^4}{z^6}}}{{x{y^3}{z^3}\left( {{x^2}{y^4}{z^6} + 1} \right)}} \cr & {f_y}\left( {x,y,z} \right) = - \frac{{2xy{z^3}}}{{\left( {{x^2}{y^4}{z^6} + 1} \right)}} \cr & \cr & {\text{Calculate }}{f_z}\left( {x,y,z} \right) \cr & {f_z}\left( {x,y,z} \right) = \frac{\partial }{{\partial z}}\left[ {{{\tan }^{ - 1}}\left( {\frac{1}{{x{y^2}{z^3}}}} \right)} \right] \cr & {f_z}\left( {x,y,z} \right) = \frac{{\frac{\partial }{{\partial z}}\left[ {\frac{1}{{x{y^2}{z^3}}}} \right]}}{{1 + {{\left( {1/x{y^2}{z^3}} \right)}^2}}} \cr & {f_z}\left( {x,y,z} \right) = \frac{{ - \frac{3}{{x{y^2}{z^4}}}}}{{1 + 1/{x^2}{y^4}{z^6}}} \cr & {f_z}\left( {x,y,z} \right) = \frac{{ - \frac{3}{{x{y^2}{z^4}}}}}{{\left( {{x^2}{y^4}{z^6} + 1} \right)/{x^2}{y^4}{z^6}}} \cr & {f_z}\left( {x,y,z} \right) = - \frac{{3{x^2}{y^4}{z^6}}}{{x{y^2}{z^4}\left( {{x^2}{y^4}{z^6} + 1} \right)}} \cr & {f_z}\left( {x,y,z} \right) = - \frac{{3x{y^2}{z^2}}}{{\left( {{x^2}{y^4}{z^6} + 1} \right)}} \cr} $$