Let $f ( x , y ) = \frac { 1 } { x + y ^ { 2 } }$ . Find the gradient vector $\nabla f ( 1,1 )$ at the point $( x , y ) = ( 1,1 )$ .
A) $- i - j$

B) $- \frac { 1 } { 4 } \mathbf { i } - \frac { 1 } { 2 } \mathbf { j }$
C) $- \mathbf { i } - \frac { 1 } { 2 } \mathbf { j }$
D) $- \frac { 1 } { 2 } \mathbf { i } - \mathbf { j }$
E) $\mathbf { i } + \mathbf { j }$
F) $\frac { 1 } { 4 } \mathbf { i } + \frac { 1 } { 2 } \mathbf { j }$

G) $\mathbf { i } + \frac { 1 } { 2 } \mathbf { j }$

H) $\frac { 1 } { 2 } \mathbf { i } + \mathbf { j }$

Question

Let $f ( x , y ) = \frac { 1 } { x + y ^ { 2 } }$ . Find the gradient vector $\nabla f ( 1,1 )$ at the point $( x , y ) = ( 1,1 )$ .
A) $- i - j$ 
 
B) $- \frac { 1 } { 4 } \mathbf { i } - \frac { 1 } { 2 } \mathbf { j }$ 
C) $- \mathbf { i } - \frac { 1 } { 2 } \mathbf { j }$ 
D) $- \frac { 1 } { 2 } \mathbf { i } - \mathbf { j }$ 
E) $\mathbf { i } + \mathbf { j }$
F) $\frac { 1 } { 4 } \mathbf { i } + \frac { 1 } { 2 } \mathbf { j }$

G) $\mathbf { i } + \frac { 1 } { 2 } \mathbf { j }$

H) $\frac { 1 } { 2 } \mathbf { i } + \mathbf { j }$

Quizplus · Accepted Answer

The gradient vector $\nabla f$ of a function $f(x, y)$ is given by $\left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right)$. For $f(x, y) = \frac{1}{x + y^2}$, $\frac{\partial f}{\partial x} = -\frac{1}{(x + y^2)^2}$ and $\frac{\partial f}{\partial y} = -\frac{2y}{(x + y^2)^2}$. Substituting $x = 1$ and $y = 1$ gives $\frac{\partial f}{\partial x} = -\frac{1}{4}$ and $\frac{\partial f}{\partial y} = -\frac{1}{2}$, thus the gradient vector at $(1,1)$ is $-\frac{1}{4}\mathbf{i} - \frac{1}{2}\mathbf{j}$.

Let f(x,y)=1x+y2f ( x , y ) = \frac { 1 } { x + y ^ { 2 } }f(x,y)=x+y21​

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