Using implicit differentiation on $y = \cos ( x + y ) , y ^ { \prime }$ is

A) $\frac { \sin ( x + y ) } { 1 + \sin ( x + y ) }$
B) $\frac { \sin ( x + y ) } { 1 - \sin ( x + y ) }$
C) $\frac { \sin ( x + y ) } { \sin ( x + y ) - 1 }$
D)-1
E) $- \frac { \sin ( x + y ) } { 1 + \sin ( x + y ) }$

Question

Using implicit differentiation on $y = \cos ( x + y ) , y ^ { \prime }$ is

A) $\frac { \sin ( x + y ) } { 1 + \sin ( x + y ) }$
B) $\frac { \sin ( x + y ) } { 1 - \sin ( x + y ) }$
C) $\frac { \sin ( x + y ) } { \sin ( x + y ) - 1 }$
D)-1
E) $- \frac { \sin ( x + y ) } { 1 + \sin ( x + y ) }$

Quizplus · Accepted Answer

Differentiating both sides of $y = \cos(x + y)$ with respect to $x$ using the chain rule gives $y' = -\sin(x + y) \cdot (1 + y')$. Solving for $y'$ yields $y' = -\frac{\sin(x + y)}{1 + \sin(x + y)}$.

Using Implicit Differentiation On y=cos⁡(x+y),y′y = \cos ( x + y ) , y ^ { \prime }y=cos(x+y),y′

Using Implicit Differentiation On $y = \cos ( x + y ) , y ^ { \prime }$