Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ using implicit differentiation. $y \ln x + y = 2$
A) $- \frac { } { x ( \ln x + 1 ) }$
B) $- \frac { x } { y ( \ln y + 1 ) }$
C) $x ( \ln x + 1 )$
D) $- \frac { 1 } { x ( \ln x + 1 ) }$
E) $\overline { x \ln x }$

Question

Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ using implicit differentiation.  $y \ln x + y = 2$  
A) $- \frac { } { x ( \ln x + 1 ) }$
B)   $- \frac { x } { y ( \ln y + 1 ) }$
C)  $x ( \ln x + 1 )$
D)  $- \frac { 1 } { x ( \ln x + 1 ) }$
E)   $\overline { x \ln x }$

Quizplus · Accepted Answer

Explanation:To find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ using implicit differentiation, we differentiate both sides of the equation $y \ln x + y = 2$ with respect to $x$.Differentiating the left-hand side, we get:$\frac { \mathrm { d } } { \mathrm {~d} x } (y \ln x + y) = \frac { \mathrm { d } } { \mathrm {~d} x } (y) \ln x + y \frac { \mathrm { d } } { \mathrm {~d} x } (\ln x) + \frac { \mathrm { d } } { \mathrm {~d} x } (y)$Simplifying, we get:$\ln x \frac { \mathrm { d } y } { \mathrm {~d} x } + y \frac { 1 } { x } + \frac { \mathrm { d } y } { \mathrm {~d} x }$Differentiating the right-hand side, we get:$\frac { \mathrm { d } } { \mathrm {~d} x } (2) = 0$Setting the two derivatives equal to each other, we get:$\ln x \frac { \mathrm { d } y } { \mathrm {~d} x } + y \frac { 1 } { x } + \frac { \mathrm { d } y } { \mathrm {~d} x } = 0$Solving for $\frac { \mathrm { d } y } { \mathrm {~d} x }$, we get:$\frac { \mathrm { d } y } { \mathrm {~d} x } = - \frac { y } { x ( \ln x + 1 ) }$Therefore, the correct answer is A.

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