Find the differential of $z = \frac { 1 } { x + y ^ { 2 } }$ at the point $( x , y ) = ( 1,1 )$ .
A) $- d x - d y$

B) $- \frac { 1 } { 4 } d x - \frac { 1 } { 2 } d y$
C)$- d x - \frac { 1 } { 2 } d y$
D)$- \frac { 1 } { 2 } d x - d y$
E) $d x + d y$
F) $\frac { 1 } { 4 } d x + \frac { 1 } { 2 } d y$

G) $d x + \frac { 1 } { 2 } d y$

H) $\frac { 1 } { 2 } d x + d y$

Question

Find the differential of $z = \frac { 1 } { x + y ^ { 2 } }$ at the point $( x , y ) = ( 1,1 )$ .
A) $- d x - d y$

B) $- \frac { 1 } { 4 } d x - \frac { 1 } { 2 } d y$ 
C)$- d x - \frac { 1 } { 2 } d y$ 
D)$- \frac { 1 } { 2 } d x - d y$ 
E) $d x + d y$ 
F) $\frac { 1 } { 4 } d x + \frac { 1 } { 2 } d y$

G) $d x + \frac { 1 } { 2 } d y$

H) $\frac { 1 } { 2 } d x + d y$

Quizplus · Accepted Answer

To find the differential $dz$ of $z = \frac{1}{x + y^2}$, we use partial derivatives. The partial derivative with respect to $x$ is $\frac{\partial z}{\partial x} = -\frac{1}{(x + y^2)^2}$, and with respect to $y$ is $\frac{\partial z}{\partial y} = -\frac{2y}{(x + y^2)^2}$. At the point $(1,1)$, these become $\frac{\partial z}{\partial x} = -\frac{1}{4}$ and $\frac{\partial z}{\partial y} = -\frac{1}{2}$. Thus, $dz = -\frac{1}{4}dx - \frac{1}{2}dy$.

Find the Differential Of z=1x+y2z = \frac { 1 } { x + y ^ { 2 } }z=x+y21​

Find the Differential Of $z = \frac { 1 } { x + y ^ { 2 } }$