Let $f ( x , y , z ) = \ln ( x y z )$ . Its gradient vector field is

A) $\nabla f ( x , y ) = \frac { 1 } { x } \mathbf { i } - \frac { 1 } { y } \mathbf { j } + \frac { 1 } { z } \mathbf { k }$
B) $\nabla f ( x , y ) = \frac { 1 } { x } \mathbf { i } + \frac { 1 } { y } \mathbf { j } + \frac { 1 } { z } \mathbf { k }$
C) $\nabla f ( x , y ) = \frac { 1 } { x } \mathbf { i } + \frac { 1 } { y } \mathbf { j } - \frac { 1 } { z } \mathbf { k }$
D) $\nabla f ( x , y ) = \frac { 1 } { x } \mathbf { i } - \frac { 1 } { y } \mathbf { j } - \frac { 1 } { z } \mathbf { k }$
E) $\nabla f ( x , y ) = - \frac { 1 } { x } \mathbf { i } + \frac { 1 } { y } \mathbf { j } + \frac { 1 } { z } \mathbf { k }$

Question

Let $f ( x , y , z ) = \ln ( x y z )$ . Its gradient vector field is

A) $\nabla f ( x , y ) = \frac { 1 } { x } \mathbf { i } - \frac { 1 } { y } \mathbf { j } + \frac { 1 } { z } \mathbf { k }$
B) $\nabla f ( x , y ) = \frac { 1 } { x } \mathbf { i } + \frac { 1 } { y } \mathbf { j } + \frac { 1 } { z } \mathbf { k }$
C) $\nabla f ( x , y ) = \frac { 1 } { x } \mathbf { i } + \frac { 1 } { y } \mathbf { j } - \frac { 1 } { z } \mathbf { k }$
D) $\nabla f ( x , y ) = \frac { 1 } { x } \mathbf { i } - \frac { 1 } { y } \mathbf { j } - \frac { 1 } { z } \mathbf { k }$
E) $\nabla f ( x , y ) = - \frac { 1 } { x } \mathbf { i } + \frac { 1 } { y } \mathbf { j } + \frac { 1 } { z } \mathbf { k }$

Quizplus · Accepted Answer

The gradient of $f(x, y, z) = \ln(xyz)$ is calculated by taking the partial derivatives with respect to each variable. The partial derivative with respect to $x$ is $\frac{1}{x}$, with respect to $y$ is $\frac{1}{y}$, and with respect to $z$ is $\frac{1}{z}$, resulting in the gradient vector $\nabla f(x, y, z) = \frac{1}{x}\mathbf{i} + \frac{1}{y}\mathbf{j} + \frac{1}{z}\mathbf{k}$.

Let f(x,y,z)=ln⁡(xyz)f ( x , y , z ) = \ln ( x y z )f(x,y,z)=ln(xyz)

Let $f ( x , y , z ) = \ln ( x y z )$