Calculate $\frac { \mathrm { d } y } { \mathrm {~d} x }$ . You need not expand your answer. $y = \frac { ( x + 2 ) ( x + 1 ) } { 2 x - 4 }$
A) $\frac { 2 x + 3 } { ( 2 x + 4 ) ^ { 2 } }$
B) $\frac { ( 2 x + 2 ) ( 2 x - 4 ) + 2 \left( x ^ { 2 } + 3 x + 2 \right) } { ( 2 x - 4 ) ^ { 2 } }$
C) $\frac { 2 x + 3 } { 2 }$
D) $\frac { ( 2 x + 3 ) ( 2 x - 4 ) - 2 \left( x ^ { 2 } + 3 x + 2 \right) } { ( 2 x - 4 ) ^ { 2 } }$
E) $\frac { ( 2 x + 2 ) ( 2 x - 4 ) + 2 \left( x ^ { 2 } + 3 x + 2 \right) } { ( 2 x - 4 ) }$

Question

Calculate $\frac { \mathrm { d } y } { \mathrm {~d} x }$ . You need not expand your answer.   $y = \frac { ( x + 2 ) ( x + 1 ) } { 2 x - 4 }$ 
A) $\frac { 2 x + 3 } { ( 2 x + 4 ) ^ { 2 } }$
B) $\frac { ( 2 x + 2 ) ( 2 x - 4 ) + 2 \left( x ^ { 2 } + 3 x + 2 \right) } { ( 2 x - 4 ) ^ { 2 } }$
C) $\frac { 2 x + 3 } { 2 }$
D) $\frac { ( 2 x + 3 ) ( 2 x - 4 ) - 2 \left( x ^ { 2 } + 3 x + 2 \right) } { ( 2 x - 4 ) ^ { 2 } }$
E) $\frac { ( 2 x + 2 ) ( 2 x - 4 ) + 2 \left( x ^ { 2 } + 3 x + 2 \right) } { ( 2 x - 4 ) }$

Quizplus · Accepted Answer

The correct answer is obtained by applying the quotient rule for differentiation, which is $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$, where $u = (x + 2)(x + 1)$ and $v = 2x - 4$. After differentiating $u$ and $v$ with respect to $x$ and substituting into the formula, the result matches option D.

Calculate dy dx\frac { \mathrm { d } y } { \mathrm {~d} x } dxdy​

Calculate $\frac { \mathrm { d } y } { \mathrm {~d} x }$