$\text { Find the derivative of the function } \frac { \sin x } { 2 x + \cos x } \text {. Simplify your answer. }$
A)
$\frac { 2 x \cos x - \cos ^ { 2 } x - 2 \sin x + \sin ^ { 2 } x } { ( 2 x + \cos x ) ^ { 2 } }$
B)
$\frac { \cos x } { ( 2 - \sin x ) ^ { 2 } }$
C) $\frac { \cos x } { 2 - \sin x }$
D) $\frac { 1 + 2 x \cos x - 2 \sin x } { ( 2 x + \cos x ) ^ { 2 } }$
E) $\frac { 2 x \cos x + \cos ^ { 2 } x - 2 \sin x - \sin ^ { 2 } x } { ( 2 x + \cos x ) ^ { 2 } }$

Question

$\text { Find the derivative of the function } \frac { \sin x } { 2 x + \cos x } \text {. Simplify your answer. }$ 
A)
$\frac { 2 x \cos x - \cos ^ { 2 } x - 2 \sin x + \sin ^ { 2 } x } { ( 2 x + \cos x ) ^ { 2 } }$
B)
$\frac { \cos x } { ( 2 - \sin x ) ^ { 2 } }$
C) $\frac { \cos x } { 2 - \sin x }$
D) $\frac { 1 + 2 x \cos x - 2 \sin x } { ( 2 x + \cos x ) ^ { 2 } }$
E) $\frac { 2 x \cos x + \cos ^ { 2 } x - 2 \sin x - \sin ^ { 2 } x } { ( 2 x + \cos x ) ^ { 2 } }$

Quizplus · Accepted Answer

We use the quotient rule: 
$$\frac{d}{dx}\left(\frac{\sin x}{2x + \cos x}\right) = \frac{(2x+\cos x)\cos x - \sin x(-2\sin x)}{(2x+ \cos x)^2}$$
Simplifying the numerator we get: 
$$2x\cos x + \cos^2 x + 2\sin^2 x = 1 + 2x\cos x - 2\sin x$$
So the derivative is:
$$\frac{1 + 2x\cos x - 2\sin x}{(2x+\cos x)^2}$$ 
which matches choice D.

Find the derivative of the function sin⁡x2x+cos⁡x. Simplify your answer. \text { Find the derivative of the function } \frac { \sin x } { 2 x + \cos x } \text {. Simplify your answer. } Find the derivative of the function 2x+cosxsinx​. Simplify your answer.

$\text { Find the derivative of the function } \frac { \sin x } { 2 x + \cos x } \text {. Simplify your answer. }$