Chapter 13 - Partial Derivatives - 13.4 Differentiability, Differentials, And Local Linearity - Exercises Set 13.4 - Page 947: 19

$$dw = \frac{{yz}}{{1 + {x^2}{y^2}{z^2}}}dx + \frac{{xz}}{{1 + {x^2}{y^2}{z^2}}}dy + \frac{{xy}}{{1 + {x^2}{y^2}{z^2}}}dz$$

$$\eqalign{ & w = {\tan ^{ - 1}}\left( {xyz} \right) \cr & {\text{Let }}w = f\left( {x,y,z} \right){\text{ }} \cr & {\text{Calculate the partial derivative }}{f_x}\left( {x,y,z} \right){\text{, treating }}y{\text{ and }}z{\text{ as constants}} \cr & {\text{ }}{f_x}\left( {x,y,z} \right) = \frac{\partial }{{\partial x}}\left[ {{{\tan }^{ - 1}}\left( {xyz} \right)} \right] \cr & {\text{use }}\frac{d}{{dx}}\left[ {{{\tan }^{ - 1}}u} \right] = \frac{1}{{{u^2} + 1}}\frac{{du}}{{dx}} \cr & {\text{ }}{f_x}\left( {x,y,z} \right) = \frac{1}{{1 + {{\left( {xyz} \right)}^2}}}\frac{\partial }{{\partial x}}\left[ {xyz} \right] \cr & {\text{ }}{f_x}\left( {x,y,z} \right) = \frac{1}{{1 + {x^2}{y^2}{z^2}}}\left( {yz} \right) \cr & {\text{ }}{f_x}\left( {x,y,z} \right) = \frac{{yz}}{{1 + {x^2}{y^2}{z^2}}} \cr & {\text{Calculate the partial derivative }}{f_y}\left( {x,y,z} \right){\text{, treating }}x{\text{ and }}z{\text{ as constants}} \cr & {\text{ }}{f_y}\left( {x,y,z} \right) = \frac{\partial }{{\partial y}}\left[ {{{\tan }^{ - 1}}\left( {xyz} \right)} \right] \cr & {\text{use }}\frac{d}{{dy}}\left[ {{{\tan }^{ - 1}}u} \right] = \frac{1}{{{u^2} + 1}}\frac{{du}}{{dy}} \cr & {\text{ }}{f_y}\left( {x,y,z} \right) = \frac{1}{{1 + {{\left( {xyz} \right)}^2}}}\frac{\partial }{{\partial y}}\left[ {xyz} \right] \cr & {\text{ }}{f_y}\left( {x,y,z} \right) = \frac{1}{{1 + {x^2}{y^2}{z^2}}}\left( {xz} \right) \cr & {\text{ }}{f_y}\left( {x,y,z} \right) = \frac{{xz}}{{1 + {x^2}{y^2}{z^2}}} \cr & {\text{Calculate the partial derivative }}{f_z}\left( {x,y,z} \right){\text{, treating }}x{\text{ and }}y{\text{ as constants}} \cr & {f_z}\left( {x,y,z} \right) = \frac{\partial }{{\partial z}}\left[ {{{\tan }^{ - 1}}\left( {xyz} \right)} \right] \cr & {\text{use }}\frac{d}{{dz}}\left[ {{{\tan }^{ - 1}}u} \right] = \frac{1}{{{u^2} + 1}}\frac{{du}}{{dz}} \cr & {\text{ }}{f_z}\left( {x,y,z} \right) = \frac{1}{{1 + {{\left( {xyz} \right)}^2}}}\frac{\partial }{{\partial z}}\left[ {xyz} \right] \cr & {\text{ }}{f_z}\left( {x,y,z} \right) = \frac{1}{{1 + {x^2}{y^2}{z^2}}}\left( {xy} \right) \cr & {\text{ }}{f_z}\left( {x,y,z} \right) = \frac{{xy}}{{1 + {x^2}{y^2}{z^2}}} \cr & \cr & {\text{The total differential of }}z{\text{ is given by}}:\,\,\left( {{\text{See page 944}}} \right) \cr & dw = {f_x}\left( {x,y,z} \right)dx + {f_y}\left( {x,y,z} \right)dy + {f_z}\left( {x,y,z} \right)dz \cr & {\text{Substitute the partial derivatives}} \cr & dw = \frac{{yz}}{{1 + {x^2}{y^2}{z^2}}}dx + \frac{{xz}}{{1 + {x^2}{y^2}{z^2}}}dy + \frac{{xy}}{{1 + {x^2}{y^2}{z^2}}}dz \cr} $$