Chapter 13 - Partial Derivatives - 13.6 Directional Derivatives And Gradients - Exercises Set 13.6 - Page 969: 45

$$\nabla f\left( { - 3,4,0} \right) = 4{\bf{i}} + 4{\bf{j}} + 4{\bf{k}}$$

$$\eqalign{ & f\left( {x,y,z} \right) = y\ln \left( {x + y + z} \right);\,\,\,\,\,\left( { - 3,4,0} \right) \cr & \cr & {\text{calculate the partial derivatives }}{f_x}\left( {x,y,z} \right){\text{,}}\,{\text{ }}{f_y}\left( {x,y,z} \right)\,\,{\text{ and }}{f_z}\left( {x,y} \right) \cr & {f_x}\left( {x,y,z} \right) = \frac{\partial }{{\partial x}}\left[ {y\ln \left( {x + y + z} \right)} \right] \cr & {\text{treat }}y{\text{ and }}z{\text{ as constants}} \cr & {f_x}\left( {x,y,z} \right) = y\frac{\partial }{{\partial x}}\left[ {\ln \left( {x + y + z} \right)} \right] \cr & {f_x}\left( {x,y,z} \right) = y\left( {\frac{1}{{x + y + z}}} \right) \cr & {f_x}\left( {x,y,z} \right) = \frac{y}{{x + y + z}} \cr & \cr & {f_y}\left( {x,y,z} \right) = \frac{\partial }{{\partial y}}\left[ {y\ln \left( {x + y + z} \right)} \right] \cr & {\text{treat }}x{\text{ and }}z{\text{ as a constant and use product rule}} \cr & {f_y}\left( {x,y,z} \right) = y\frac{\partial }{{\partial y}}\left[ {\ln \left( {x + y + z} \right)} \right] + \ln \left( {x + y + z} \right)\frac{\partial }{{\partial y}}\left[ y \right] \cr & {f_y}\left( {x,y,z} \right) = y\left( {\frac{1}{{x + y + z}}} \right) + \ln \left( {x + y + z} \right)\left( 1 \right) \cr & {f_y}\left( {x,y,z} \right) = \frac{y}{{x + y + z}} + \ln \left( {x + y + z} \right) \cr & and \cr & {f_z}\left( {x,y,z} \right) = \frac{\partial }{{\partial z}}\left[ {y\ln \left( {x + y + z} \right)} \right] \cr & {\text{treat }}x{\text{ and }}z{\text{ as a constant and use product rule}} \cr & {f_z}\left( {x,y,z} \right) = \frac{\partial }{{\partial z}}\left[ {y\ln \left( {x + y + z} \right)} \right] \cr & {\text{treat }}y{\text{ and }}z{\text{ as constants}} \cr & {f_z}\left( {x,y,z} \right) = y\frac{\partial }{{\partial z}}\left[ {\ln \left( {x + y + z} \right)} \right] \cr & {f_z}\left( {x,y,z} \right) = y\left( {\frac{1}{{x + y + z}}} \right) \cr & {f_z}\left( {x,y,z} \right) = \frac{y}{{x + y + z}} \cr & \cr & {\text{The gradient of the function }}f\left( {x,y,z} \right){\text{ is defined by }}\left( {{\text{see page 963}}} \right) \cr & \nabla f\left( {x,y,z} \right) = {f_x}\left( {x,y,z} \right){\bf{i}} + {f_y}\left( {x,y,z} \right){\bf{j}} + {f_z}\left( {x,y,z} \right){\bf{k}} \cr & {\text{substituting the partial derivatives, we obtain}} \cr & \nabla f\left( {x,y,z} \right) = \frac{y}{{x + y + z}}{\bf{i}} + \left[ {\frac{y}{{x + y + z}} + \ln \left( {x + y + z} \right)} \right]{\bf{j}} + \frac{y}{{x + y + z}}{\bf{k}} \cr & \cr & {\text{the gradient of }}f{\text{ at }}\left( { - 3,4,0} \right){\text{ is}} \cr & \nabla f\left( { - 3,4,0} \right) = \frac{4}{{ - 3 + 4 + 0}}{\bf{i}} + \left[ {\frac{4}{{ - 3 + 4 + 0}} + \ln \left( { - 3 + 4 + 0} \right)} \right]{\bf{j}} + \frac{4}{{ - 3 + 4 + 0}}{\bf{k}} \cr & \nabla f\left( { - 3,4,0} \right) = \frac{4}{{ - 3 + 4 + 0}}{\bf{i}} + \left[ {\frac{4}{{ - 3 + 4 + 0}} + \ln \left( { - 3 + 4 + 0} \right)} \right]{\bf{j}} + \frac{4}{{ - 3 + 4 + 0}}{\bf{k}} \cr & \nabla f\left( { - 3,4,0} \right) = 4{\bf{i}} + 4{\bf{j}} + 4{\bf{k}} \cr} $$