Let $f ( x , y , z ) = \left( x ^ { 2 } + y ^ { 2 } + z ^ { 2 } \right) ^ { - \frac { 1 } { 2 } }$ .Its gradient vector field is

A) $\nabla f ( x , y , z ) = - \frac { x \mathbf { i } - y \mathbf { j } + z \mathbf { k } } { \left( x ^ { 2 } + y ^ { 2 } + z ^ { 2 } \right) ^ { \frac { 3 } { 2 } } }$
B) $\nabla f ( x , y , z ) = \frac { x \mathbf { i } + y \mathbf { j } + z \mathbf { k } } { \left( x ^ { 2 } + y ^ { 2 } + z ^ { 2 } \right) ^ { \frac { 3 } { 2 } } }$
C) $\nabla f ( x , y , z ) = - \frac { x \mathbf { i } + y \mathbf { j } + z \mathbf { k } } { \left( x ^ { 2 } + y ^ { 2 } + z ^ { 2 } \right) ^ { \frac { 3 } { 2 } } }$
D) $\nabla f ( x , y , z ) = \frac { x \mathbf { i } - y \mathbf { j } + z \mathbf { k } } { \left( x ^ { 2 } + y ^ { 2 } + z ^ { 2 } \right) ^ { \frac { 3 } { 2 } } }$
E) $\nabla f ( x , y , z ) = - \frac { x \mathbf { i } - y \mathbf { j } - z \mathbf { k } } { \left( x ^ { 2 } + y ^ { 2 } + z ^ { 2 } \right) ^ { \frac { 3 } { 2 } } }$

Question

Let $f ( x , y , z ) = \left( x ^ { 2 } + y ^ { 2 } + z ^ { 2 } \right) ^ { - \frac { 1 } { 2 } }$ .Its gradient vector field is

A) $\nabla f ( x , y , z ) = - \frac { x \mathbf { i } - y \mathbf { j } + z \mathbf { k } } { \left( x ^ { 2 } + y ^ { 2 } + z ^ { 2 } \right) ^ { \frac { 3 } { 2 } } }$
B) $\nabla f ( x , y , z ) = \frac { x \mathbf { i } + y \mathbf { j } + z \mathbf { k } } { \left( x ^ { 2 } + y ^ { 2 } + z ^ { 2 } \right) ^ { \frac { 3 } { 2 } } }$
C) $\nabla f ( x , y , z ) = - \frac { x \mathbf { i } + y \mathbf { j } + z \mathbf { k } } { \left( x ^ { 2 } + y ^ { 2 } + z ^ { 2 } \right) ^ { \frac { 3 } { 2 } } }$
D) $\nabla f ( x , y , z ) = \frac { x \mathbf { i } - y \mathbf { j } + z \mathbf { k } } { \left( x ^ { 2 } + y ^ { 2 } + z ^ { 2 } \right) ^ { \frac { 3 } { 2 } } }$
E) $\nabla f ( x , y , z ) = - \frac { x \mathbf { i } - y \mathbf { j } - z \mathbf { k } } { \left( x ^ { 2 } + y ^ { 2 } + z ^ { 2 } \right) ^ { \frac { 3 } { 2 } } }$

Quizplus · Accepted Answer

The gradient of $f(x, y, z)$ involves taking the partial derivatives with respect to $x$, $y$, and $z$, and then combining these derivatives into a vector. The derivative of $(x^2 + y^2 + z^2)^{-1/2}$ with respect to each variable involves applying the chain rule, resulting in a factor of $-1/2$ from the exponent and another factor from the derivative of the inside function (e.g., $2x$ for $x$), which simplifies to $-x$, $-y$, and $-z$ for each respective partial derivative. These are then divided by the cube of the original function due to the power rule, leading to the correct answer being the vector with components $-x$, $-y$, and $-z$ over $(x^2 + y^2 + z^2)^{3/2}$, which matches choice C.

Let f(x,y,z)=(x2+y2+z2)−12f ( x , y , z ) = \left( x ^ { 2 } + y ^ { 2 } + z ^ { 2 } \right) ^ { - \frac { 1 } { 2 } }f(x,y,z)=(x2+y2+z2)−21​

Let $f ( x , y , z ) = \left( x ^ { 2 } + y ^ { 2 } + z ^ { 2 } \right) ^ { - \frac { 1 } { 2 } }$